The definition of $ A^{\mu}$ that I've been given is as an object created from the scalar field $\phi$ and the vector field $\vec{A}$ i.e:
$$ A^{\mu} = (\phi,\frac{\vec{A}}{c}) $$ [in cgs units]
How do I formally prove that this is in fact a valid four-vector, and not some jumble of objects?
(The Proof that four-potential is a four-vector question does not answer this - it posits some strange specific case.)
To see that this is a four-vector we must first understand where it comes from. We have the Maxwell equations, where I will use $\dot X = \partial_w X = c^{-1} \partial_t X$ to write as, $$\begin{align} \nabla\cdot E &= 4\pi\rho& \nabla\times E&= - \dot B\\ \nabla\cdot B &= 0 &\nabla\times B&= 4\pi c^{-1} J + \dot E, \end{align}$$ and after using $\nabla\cdot B = 0$ to define a family of $A$ via $B = \nabla\times A$ and the consequent $\nabla \times (E + \dot A) = 0$ to define a family of $\varphi$ via $E = -\dot A - \nabla \varphi,$ we can derive that the remaining two equations state,$$ \begin{align} -\nabla\cdot\dot A - \ddot\varphi &= 4\pi\rho\\ \nabla(\nabla\cdot A) - \nabla^2 A &= 4\pi c^{-1}J - \ddot A - \nabla\dot\varphi \end{align} $$ which can be handily rewritten (by defining $\Box X =\ddot X -\nabla^2 X,$ and $\lambda = \dot\varphi + \nabla\cdot A$) as,$$ \begin{align} \Box\varphi &= 4\pi\rho + \dot \lambda,\\ \Box A &= 4\pi c^{-1}J - \nabla\lambda. \end{align}$$ Now we could have chosen a different set of potentials and still obtained the same fields: we know that we can add any $\nabla\chi$ to $A$ and preserve $B=\nabla\times A$ because the curl of a gradient is zero; looking at what's needed to preserve $E$ tells us only that we should simultaneously subtract $\dot\chi$ from $\varphi$ to preserve $E$. This does not affect the form of the last two equations directly, but it does map the value of $\lambda\mapsto \lambda - \Box \chi,$ and so we can essentially choose a different functional form for $\lambda,$ on the grounds that if we want, say, $\lambda = \dot\varphi$ (Coulomb gauge) we can take any other solution $(A, \varphi)$ and calculate its $\lambda$ and then solve $\Box\chi = \lambda - \dot\varphi$ to find a $\chi$ which gives us a different set of fields which has this functional form.
Of course the Lorenz gauge that we now assume is to solve for $A,\varphi$ such that $\lambda=0,$ yielding,$$ \begin{align} \Box\varphi &= 4\pi \rho\\ \Box A_x &= 4\pi c^{-1}J_x\\ \Box A_y &= 4\pi c^{-1}J_y\\ \Box A_z &= 4\pi c^{-1}J_z \end{align}$$
I am going through all of this for three reasons:
The rough proof that $(\rho, J/c)$ is a four-vector involves imagining a static charge distribution first: a static charge distribution $(\rho_0, 0)$ becomes $(\gamma~\rho_0, -\gamma~ \rho_0 \vec \beta)$ under a Lorentz transform; this is easily seen to be a $(\rho, J/c)$ pair. But if this is a four-vector then a more complex arrangement is a four-vector, precisely because an arbitrary charge-current distribution can be arbitrarily-well approximated by a superposition of a bunch of little static charge-current distributions that have been boosted in various ways. If all of them individually transform appropriately over Lorentz transforms, then their sum must, too, because Lorentz transforms are linear.
If you agree with me that $(\rho, J/c)$ is a four-vector then this must also be the case for $(\varphi, A/c)$, also by a linearity argument, though it is a sufficiency and not a necessity (as it must be: the Lorenz gauge has not 100% nailed down the exact fields $\varphi$ and $A$, so there must be other fields that are not the Lorentz transform of these fields which also pass the Lorenz gauge and are valid).
Just examine the equations in the Lorenz gauge, after a Lorentz boost of the charge fields by $\beta$ in the $\hat z$ direction: $$\begin{align} \Box\phi' &= 4\pi ~\gamma~(\rho - \beta c^{-1}J_z)\\ \Box A_x' &= 4\pi c^{-1}~J_x\\ \Box A_y' &= 4\pi c^{-1}~J_y\\ \Box A_z' &= 4\pi c^{-1}~\gamma~(J_z - \beta c\rho) \end{align}$$It is clearly sufficient to solve these equations to find $A_x' = A_x$ and $A_y' = A_y.$ For $A_z'$ it is clearly sufficient to have $A_z' = \gamma~A_z - \gamma\beta\phi$ and for $\varphi' = \gamma~\varphi - \gamma\beta A_z.$
Thus: if you have solved $(\rho, J/c)$ for the fields $(\varphi, A)$ in the Lorenz gauge, then you can obtain a valid solution for the fields $(\varphi', A')$ that you would get from solving the same equations for the Lorentz-boosted $(\rho', J'/c)$ simply by Lorentz-transforming the fields $(\varphi, A)$ as a four-vector.
In other words: there are ways to do electromagnetism in which that pair is not a four-vector, but it does no harm to assume that it is.
I am answering this question for a homework so I'll try to share here what I am thinking:
Let us take for granted that at least $J^\mu$ is a 4-vector. Then, let's go back to Maxwell equations, which in the Lorentz gauge ($\partial_\mu A^\mu=0$) are (I'm doing GR so I'm using the metric $\eta_{\mu\nu}=diag(-1,1,1,1)$, and there are a couple of different signs with respect to the particle physics one, hopefully that should not interfere with the reasoning) \begin{equation} \Box A^\mu = -J^\mu \end{equation} If $J^\mu$ is a 4-vector, it is the quantity $\Box A^\mu$ (and not just $A^\mu$) that must behave like a 4-vector.
As already pointed out by some previous answers, as $\Box$ is a Lorentz scalar, it follows that $A^\mu$ is a 4-vector, up to a function with null Dalembertian: if $\tilde A^\mu(x)$ is the 4-vector that satisfies $\Box \tilde A^\mu=-J^\mu$, and $\chi^\mu$ is any collection of functions which satisfy $\Box \chi^\mu=0$ for all $\mu$, then the quantity \begin{equation} A^\mu = \tilde A^\mu+\chi^\mu \end{equation}satisfies Maxwell's equations and the requirement that $\Box A^\mu$ be a 4-vector. This is not yet enough to state that $A^\mu$ is a 4-vector, but the additional constraint of the Lorentz gauge imposes \begin{equation} 0=\partial_\mu A^\mu=\partial_\mu\tilde A^\mu+\partial_\mu \chi^\mu\Rightarrow \partial_\mu\tilde A^\mu=-\partial_\mu \chi^\mu \end{equation}so again, the quantity $\partial_\mu \chi^\mu$ must behave like a scalar, which means that the set of functions $\chi^\mu$ has a pure 4-vector component $\tilde\chi^\mu$, and is defined up to a function $h^\mu$, not necessarily a 4-vector, but with null divergence: \begin{equation} \chi^\mu=\tilde \chi^\mu+h^\mu, \partial_\mu h^\mu=0 \end{equation} Finally: \begin{equation} A^\mu=\underbrace{\tilde A^\mu}_{4-v}+\underbrace{\tilde \chi^\mu}_{4-v}+h^\mu \end{equation}Changing reference frame, $\tilde A^\mu$ and $\tilde \chi^\mu$ will transform as 4-vectors into $\tilde A'^\mu$ and $\tilde \chi'^\mu$, while $h^\mu$ in principle might transform in some $h'^\mu$ which is not $\Lambda^\mu\,_\nu h^\nu$, and which needs not satisfy $\partial_\mu' h'^\mu=0$. But thanks to the gauge invariance, we can always perform a redefinition of the field $A^\mu$ which cancels the nonzero 4-divergence: \begin{equation} A^\mu\rightarrow A'^\mu=\underbrace{\tilde A'^\mu}_{4-v}+\underbrace{\tilde \chi'^\mu}_{4-v}+h'^\mu+\partial'^\mu \Phi \end{equation} \begin{equation} \partial_\mu h'^\mu+\partial_\mu\partial^\mu \Phi=0 \Rightarrow \Box\Phi =-\partial_\mu h'^\mu \end{equation}The function $\Phi$ can always be determined using the Green function for the Dalembertian. Everything on the left hand side is a scalar, so $h'^\mu$ can only be a 4-vector $\tilde h'^\mu$ plus a divergenceless part: \begin{equation} h'^\mu = \tilde h'^\mu+g'^\mu,\quad\partial_\mu' g'^\mu=0 \end{equation}So the field in this new frame is \begin{equation} A'^\mu=\underbrace{\tilde A'^\mu}_{4-v}+\underbrace{\tilde \chi'^\mu}_{4-v}+\underbrace{\tilde h'^\mu}_{4-v}+\underbrace{\partial'^\mu \Phi}_{4-v} +g'^\mu \end{equation}Comparing the two potentials in the two frames: \begin{equation} \begin{array}{cccccc} A^\mu & = & \underbrace{\tilde A^\mu+\tilde \chi^\mu}_{4-v} & + & \underbrace{h^\mu}_{\partial_\mu h^\mu=0} & \\ A'^\mu & = & \underbrace{\tilde A'^\mu+\tilde \chi'^\mu+\tilde h'^\mu+\partial'^\mu \Phi}_{4-v} & + & \underbrace{g'^\mu}_{\partial'_\mu g'^\mu=0}&\\ A'^\mu & = & \underbrace{\tilde A'^\mu+\tilde \chi'^\mu}_{4-v} & + & \underbrace{g'^\mu}_{\partial'_\mu g'^\mu=0}& \Rightarrow (\text{choose }\partial'^\mu \Phi=-h'^\mu) \end{array} \end{equation}Long story short: choosing suitably $\Phi$ we could cancel $\tilde h'^\mu$ out and at least make the 4-vector part of the two fields coincide. But we'll never be able to cut the non-vector, divergenceless part, out of the field, which then will remain globally a non-4-vector.
Imho, $A^\mu$ needs to be defined to be a 4-vector.
I want to show that the four current is indeed a four vector the proof for the potentials follows from the differential equations as sufficient solution rather than necessary as answered by other user.
$x_0$ be space co-ordinates of frame where charge density $\rho_0$ is at rest
$x$ space co-ordinates of another frame where charge density is $\rho_x$ and moving away with speed $v_x$ from the rest frame
$\gamma=\frac{1}{\sqrt{1-\frac{v_x^2}{c^2}}}$
Then because of charge invariance and length contraction $\int\rho_0 d^3x_0=\int \rho_x d^3x=\frac{1}{\gamma}\int \rho_x d^3x_0$
$\rho_x=\gamma\rho_0$
current density $J_x=\rho_x v_x= \gamma\rho_0 v_x$
now the vector $(\frac{J_x}{c}, \gamma \rho_0)=\rho_0 (\gamma \frac{v_x}{c} , \gamma )$ is a four vector because $ (\gamma \frac{v_x}{c} , \gamma )$ is proper velocity in special relativity
The four-vectors are defined by transformation laws. If the physical effects of $A^\mu$ in one frame are the same as physical effects of $\Lambda^\mu_\nu A^\nu$ in different frame - then one can say that $A$ is a four-vector.
Physical effects can be seen from the Maxwell equations, which have contravariant form when expressed in terms of the Field Strength tensor:
$$\partial_\mu F^{\mu\nu} = 0$$
Transformation law of $F$ implies four-vector transformation law of $A$.
See Electromagnetic Tensor for details.
@thesundayscientist. No, the Maxwell 4-potential $A^{\mu}$ is not a valid 4-vector field (i.e. transforming under the fundamental representation of $\mathcal{Lor} \left(3,1\right)$, or equivalently, under the $\left(1/2,1/2\right)$ representation of $\mbox{SL}(2,\mathbb{C})$), at least not in quantum field theory. The gauge symmetry spoils the correct transformation law under an element $\Lambda \in \mathcal{Lor} \left(3,1\right)$ of the quantized 4-potential field, as shown by S.Weinberg on pages 249-251 of the 1st volume of his famous treatise on QFT. However, the so-called Faraday tensor of electrodynamics $F_{\mu\nu}$ is a valid antisymmetric Lorentz tensor field, because the curl drops the gauge term. This is also spelled out by Weinberg, op.cit.
EDIT: There's a conflict of definitions here which is not addressed. A 4-vector field in classical field theory is [completely ignoring all diff.geom. aspects] a smooth assignment of a 4-tuple of functions to every point in spacetime, with the property that, under a change of inertial observers whose coordinates of their respective points are transformed by a restricted Lorentz transformation, they are transformed covariantly: $V^{\mu} (x) \longrightarrow V'^{\mu}(x') = \Lambda^{\mu}_{~\nu}V^{\nu}(x)$. Under this definition, you can say that the Maxwell 4-potential is a 4-vector.
A 4-vector field in quantum field theory is defined as per Wightman axioms as: the 4-tuple of Fock-space operator-valued distributions on the flat Minkowski spacetime whose covariance under $\mbox{ISL}(2,\mathbb{C})$ is given by 
. What Wightman is saying is that the quantized electromagnetic potential - precisely due to the gauge invariance coming from the masslessness of the classical field - cannot be covariant under $\mbox{ISL}(2,\mathbb{C})$ in the sense of eqn. (4.1.82a) of Lopuszanski's book on axiomatical QFT
.
