How do we prove that the 4-current $j^\mu$ transforms like $x^\mu$ under Lorentz transformation? Given that the position vector $\textbf{r}$ to be a vector under rotation, we mean that it transforms under rotation as $\textbf{r}^\prime=\mathbb{R}\textbf{r}$. Now, taking two time-derivatives of it, one can easily see that the acceleration $\textbf{a}=\ddot{\textbf{r}}$ transforms as $\textbf{a}^\prime=\mathbb{R}\textbf{a}$ i.e., also behaves as a vector under rotation.
Now a four-vector is something which transforms under Lorentz transformation as $x^\mu$ does. Given the transformation of $x^\mu$: $$x'^\mu=\Lambda^{\mu}{}_{\nu} x^\nu\tag{1}$$ how can one show that the four-current density $j^\mu$ also transforms like (1) preferably from the definition $j^\mu=(c\rho,\textbf{j})$?
 A: Charge density $\rho$ and current density $\mathbf j$ obey Maxwell's equations in all inertial frames. This means that in every inertial frame, the current density 4-tuple obeys the same relation; in the original frame, we have
$$
(c\rho,\mathbf j) = (c\epsilon_0\nabla\cdot \mathbf E,\nabla\times\mathbf B/\mu_0 - \epsilon_0\partial_t \mathbf E).
$$
and in the primed frame moving with respect to the first frame, we have
$$
(c\rho',\mathbf j') = (c\epsilon_0\nabla'\cdot \mathbf E',\nabla'\times\mathbf B'/\mu_0 - \epsilon_0\partial_t' \mathbf E').
$$
We can express fields $\mathbf E',\mathbf B'$ operations $\partial_t',\nabla'$ on the right-hand side withf $\mathbf E,\mathbf B$ and operations  $\partial_t,\nabla$, using the transformation formulae for fields $\mathbf E,\mathbf B$ in relativistic theory$^*$. When that is done, it can be inferred that the 4-tuple transforms as a four-vector. This method of proof is tedious but quite convincing.
$^*$Those follow from general relativistic transformation of 3-force in relativistic mechanics; see Frobenius' answer, formula 11, here:
https://physics.stackexchange.com/a/411129/31895
or the paper https://arxiv.org/abs/physics/0507099 . When applied to the Lorentz formula, which defines electric and magnetic field in every inertial frame:
$$
\mathbf F =q\mathbf E + q\mathbf v\times\mathbf B.
$$
we may derive transformation formulae for the fields.
Easier (but less convincing) way to prove $j$ is a four-vector: Maxwell's equations imply
$$
j^\mu = \partial_\nu F^{\nu\mu}.
$$
Because $F^{\nu\mu}$ is a four-tensor$^{**}$, the expression $\partial_\nu F^{\nu\mu}$ defines a four-tensor.
$^{**}$ This follows from the definition of $F$ -- antisymmetric tensor whose components are formed from components of electric and magnetic field -- and the transformation formulae for those fields mentioned above. Alternatively, if we accept that there is universal equation of motion of a test particle in EM field for every frame and every four-velocity
$$
qF^{\nu\mu}u_\mu = m\,du^\nu/d\tau
$$
it seems that $F$ must be a four-tensor. All other-than-$F$ quantities transform as four-tensors ($q,m,\tau$ are invariant, $u$ is a 4-vector by definition), so $F^{\nu\mu}u_\mu$ is a four-tensor. Then, it is plausible that $F$ in this expression is a four-tensor as well (this is the problematic part - how to make sure that F must be tensor here?).
A: $\color{blue}{\textbf{ANSWER A}}\:$ (based on charge invariance, paragraph extracted from Landau)
The answer is given in ACuriousMind's comment as pointed out also by WetSavannaAnimal aka Rod Vance. Simply I give the details copying from "The Classical Theory of Fields", L.D.Landau and E.M.Lifshitz, Fourth Revised English Edition :   

$\boldsymbol{\S}\: \textbf{28. The four-dimensional current vector}$ 
Instead of treating charges as points, for mathematical convenience we frequently consider them to be distributed continuously in space. Then we can introduce the "charge density" $\:\varrho\:$ such that $\:\varrho dV\:$ is the charge contained in the volume $\: dV$. The density $\:\varrho\:$ is in general a function of the coordinates and the time. The integral of $\:\varrho\:$ over a certain volume is the
  charge contained in that volume....... 
.......The charge on a particle is, from its very definition, an invariant quantity, that is, it does not depend on the choice of reference system. On the other hand, the density $\:\varrho\:$ is not generally an invariant--only the product $\:\varrho dV\:$ is invariant.
Multiplying the equality $\:de=\varrho dV\:$ on both sides with $\:dx^{i}\:$:
  \begin{equation}
  de\,dx^{i}=\varrho dVdx^{i}=\varrho dVdt\dfrac{dx^{i}}{dt}
\nonumber
\end{equation}
  On the left stands a four-vector (since $\:de\:$ is a scalar and $\:dx^{i}\:$ is a four-vector). This means that the right side must be a four-vector. But $\: dVdt\:$ is a scalar(1), and so $\:\varrho dx^{i}/dt\:$ is a four-vector.This vector (we denote it by $\:j^{i}$) is called the current four-vector:
  \begin{equation}
  j^{i}=\varrho \dfrac{dx^{i}}{dt}.
\tag{28.2}
\end{equation}
The space components of this vector form the current density vector,
  \begin{equation}
  \mathbf{j}=\varrho \mathbf{v},
\tag{28.3}
\end{equation}
  where $\:\mathbf{v}\:$ is the velocity of the charge at the given point. The time component of the four vector (28.2) is $\:c\varrho$. Thus
  \begin{equation}
  j^{i}=\left(c\varrho ,\mathbf{j}\right)
\tag{28.4}
\end{equation} 


(1)
Note by Frobenius : We have
\begin{equation}
  dVd(ct)=dx^{1}dx^{2}dx^{3}dx^{4}
\tag{01}
\end{equation}
Now, for the relation between the infinitesimal 4-volumes in Minkowski space
\begin{equation}
 dx'^{1}dx'^{2}dx'^{3}dx'^{4} =\begin{vmatrix}
  \dfrac{\partial x'_{1}}{\partial x_{1}}& \dfrac{\partial x'_{1}}{\partial x_{2}}&\dfrac{\partial x'_{1}}{\partial x_{3}}&\dfrac{\partial x'_{1}}{\partial x_{4}}\\
 \dfrac{\partial x'_{2}}{\partial x_{1}}& \dfrac{\partial x'_{2}}{\partial x_{2}}&\dfrac{\partial x'_{2}}{\partial x_{3}}&\dfrac{\partial x'_{2}}{\partial x_{4}}\\
  \dfrac{\partial x'_{3}}{\partial x_{1}}& \dfrac{\partial x'_{3}}{\partial x_{2}}&\dfrac{\partial x'_{3}}{\partial x_{3}}&\dfrac{\partial x'_{3}}{\partial x_{4}}\\
\dfrac{\partial x'_{4}}{\partial x_{1}}& \dfrac{\partial x'_{4}}{\partial x_{2}}&\dfrac{\partial x'_{4}}{\partial x_{3}}&\dfrac{\partial x'_{4}}{\partial x_{4}} \end{vmatrix}
dx^{1}dx^{2}dx^{3}dx^{4}=\left\vert\dfrac{\partial\left(x'^{1},x'^{2},x'^{3},x'^{4}\right)}{\partial\left(x^{1},x^{2},x^{3},x^{4}\right)}\right\vert dx^{1}dx^{2}dx^{3}dx^{4}
\tag{02}
\end{equation}
where $\:\left\vert\partial\left(x'^{1},x'^{2},x'^{3},x'^{4}\right)/\partial\left(x^{1},x^{2},x^{3},x^{4}\right)\right\vert\:$ the Jacobian, that is determinant of the Jacobi matrix. But the Jacobi matrix is the Lorentz matrix $\:\Lambda\:$ with $\:\det(\Lambda)=+1$,
that is 
\begin{equation}
 \left\vert\dfrac{\partial\left(x'^{1},x'^{2},x'^{3},x'^{4}\right)}{\partial\left(x^{1},x^{2},x^{3},x^{4}\right)}\right\vert=\det(\Lambda)=+1 
\tag{03}
\end{equation}
so
\begin{equation}
 dx'^{1}dx'^{2}dx'^{3}dx'^{4} =dx^{1}dx^{2}dx^{3}dx^{4}=\text{scalar invariant}
\tag{04}
\end{equation}

A: I think the starting point of this is to see how $j^\mu$ is defined. In the absence of charges the EM action is given by
$$
S= \int d^4 x F_{\mu \nu} F^{\mu \nu}
$$
where $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ which comes from gauge invariance. The equation of motion is
$$
\partial_\mu F^{\mu \nu} =0
$$
and introducing charges means that by Lorentz covariance the only possibility is
$$
\partial_\mu F^{\mu \nu} = j^\nu
$$
Then writing everything explicitly in terms of electrico-magnetic fields, charges and currents would give the desired relation. I think one ambiguity would be in $A_\mu = ( \pm \Phi,\vec A)$ and a choice would have to be made and as the Lagrangian has $A_\mu j^\mu$. Here one would have to invoke some physical idea like Prahar mentioned above.
A: You can take charge conservation as your starting point. This can be written as:
$$ \frac{\partial\rho}{\partial t} = \partial_{i} j^i = \nabla\cdot \vec{J} $$
Since this is an experimental fact, it is a good starting point. The above equation can now be re-written in a "more" co-variant formulation as:
$$ \partial_\mu j^\mu = 0 $$
From this equation you can clearly deduce that $j^\mu$ must transform like $x^\mu$. 
A: Rather than approaching from the fields ($F^{\mu\nu}$, $A^\mu$, etc.), a more direct approach, starting from matter, can be suggested.
In fact, the charge density $\rho (t, x^i )$ and the current density $J^i (t, x^i )$ for a point charge $q$ a charge moving with velocity $V^i (t) = \frac{d}{dt} w^i (t) $ is
$$ \rho (t, x^i) = q \delta^{(3)}(x^i - w^i(t)) $$
$$ J^i (t,x^i) = q V^i (t) \delta^{(3)}(x^i - w^i(t))  $$
and we can combine these and write as
$$ J^\mu (t, x^i) = q \left( 1, V^i (t) \right) \delta^{(3)}(x^i - w^i(t)), $$
where $\mu = 0, \ 1, \ 2, \ 3$ and $ i = 1, \ 2,\ 3 $.
Now, please observe that, if we reparametrize the particle's space-time position by the proper time ($t = t(\tau) := w^0 (\tau)$ and $w^i = w^i(\tau)$),
$$ J^\mu (x^\mu) = q \int d \tau \ u^\mu (\tau) \delta^{(4)}(x^\mu - w^\mu(\tau))  \cdots (\ast)$$
$$ \left(  \delta^{(4)}(x^\mu - w^\mu(\tau)) = \delta(t - w^0(\tau) ) \delta^{(3)}(x^i - w^i(\tau)) \right),$$
where $\tau$ and $u^\mu = \frac{d}{d\tau} w^\mu = \frac{dt}{d\tau} ( 1, V^i )$ are the proper time and 4-velocity of the point charge, respectively.
(This equation is introduced not only in relativity texts but also in books regarding electromagnetism (Jackson Ch.12, for example).)
Please notice that from this expression, we can obviously see that $J^\mu$ transforms like $u^\mu$ which is a contravariant quantity ($u^\mu = dx^\mu/d\tau$ and $dx^\mu$ is by definition contravariant and $d\tau$ is Lorentz invariant). This can be the answer of your question. Physically (or geometrically), equation $(\ast)$ provides a picture of "the distribution of charge and current for a charged particle as a superposition of charges that momentarily flash into existence and then flash out of existence." (Misner, Thorne, Wheeler: 120-121) 4-current is just a flow of "electromagnetic existence," so it is plausible that $J^\mu$ follows the transformation properties of $u^\mu$.
For continuous distributions, we just drop the integral and the delta function in equation $(\ast)$ and "continuous-ize" it:
$$ J^\mu = \varrho u^\mu   ,$$
where $\varrho$ is the Lorentz invariant charge density ("continuous-ized $q$")-the charge density seen as in the (momentarily co-moving) rest frame.
So evidently, $J^\mu$ is just a multiple of $u^\mu$, which is a contravariant quantity. Thus, $J^\mu$ is contravariant, i.e. "transforms like $dx^\mu$ under Lorentz transformation."
