Is the magnetic field really a relativistic effect of the electric field? Many sources have claimed that the magnetic field is a relativistic effect of the electric field. In that case, an object must have a non-zero charge and velocity to experience the magnetic field, to get the effect of length contraction. However, experiments have shown that current flowing through a wire can deflect compasses. However, those compasses were stationary, and thus are in the same frame of reference as the wire, and should not experience length contraction and not experience the repulsion/attraction by the magnetic field right?
 A: TL;DR - No, that's not correct. Any relativistic force in $(3+1)$-dimensions which is proportional to $\mathbf u$ can ultimately be expressed in terms of $3$-vector field $\vec E$ and $\vec B$ which mix together under boosts. Neither is more fundamental than the other, and both are just aspects of the fundamental electromagnetic field as viewed from a specific frame.

Consider starting from a (special) relativistic viewpoint using Cartesian coordinates.  Recall that the 4-velocity $\mathbf u\equiv (\gamma c, \gamma \vec v)$ of a massive particle obeys the constraint $\mathbf u \cdot \mathbf u \equiv \eta_{\mu\nu} u^\mu u^\nu = c^2$, where $\eta$ is the Minkowski metric and where we choose the mostly-minus convention (+---) for its signature. Differentiating this with respect to the proper time $\tau$ yields the condition
$$\mathbf a \cdot \mathbf u \equiv \frac{d\mathbf u}{d\tau} \cdot \mathbf u = 0$$
where $\mathbf a$ is the 4-acceleration.  This is a non-negotiable constraint on the relativistic dynamics of a massive particle.

Now let's postulate the existence of a force $f$  which is proportional to $\mathbf u$, i.e. a force of the form $f=F\mathbf u$ for some matrix $F$.  If we take the relativistic generalization of Newton's law to be $m\mathbf a = f$, then it is clear that $F$ must satisfy
$$\mathbf u \cdot F\mathbf u = \eta_{\mu\nu} u^\mu F^\nu_{\ \ \rho} u^\rho = F_{\mu\rho} u^\mu u^\rho = 0$$
for any $\mathbf u$. This implies immediately that $F_{\mu\nu}$ is antisymmetric, with its components taking the form
$$F_{\mu\nu} = \pmatrix{0&F_{01} & F_{02} & F_{03} \\ -F_{01} & 0 & F_{12} & F_{13} \\ -F_{02} & -F_{12} & 0 & F_{23} \\- F_{03} & -F_{13} & -F_{23} & 0}$$
From there we may separate out the spatial and temporal components of the force $f$ in the following way:
$$f^\mu = F^\mu_{\ \ \nu} u^\nu = \eta^{\mu\rho}F_{\rho\nu} u^\nu F_= \eta^{\nu\rho} \big(F_{\rho 0} u^0 + F_{\rho j}u^j\big)$$
$$\implies \matrix{f^0 = \gamma \sum_{j=1}^3 F_{0j} v^j\\f^i = \gamma \big(F_{0i} - \sum_{j=1}^3F_{ij} u^j\big)}$$
where $i,j$ range from $1$ to $3$ and where we've abandoned the Einstein summation convention for the time being. Note that this splitting is frame-dependent; if I choose a different reference frame, the splitting of $F$ into $F_{0i}$ and $F_{ij}$ will generically be different, in more or less the same way that the splitting of $\mathbf u$ into a temporal component and spatial components is frame-dependent.
We're almost there.  Note that in $(3+1)$-dimensions, $F_{ij}$ is an antisymmetric $3\times 3$ matrix.  It is a peculiar fact that the space of such matrices can be put into one-to-one correspondence with the space of $3$-component vectors:
$$(V^1,V^2,V^3) \leftrightarrow \pmatrix{0 & V_3 & -V_2\\-V_3 & 0 & V_1 \\ V_2 & -V_1 & 0}$$
In component form, the components of any antisymmetric matrix $\mathcal V$ can be written as $\mathcal V_{ij} = \epsilon_{ijk} V^k$ for some 3-vector $V$ (whose components can conversely be expressed as $V^k = \frac{1}{2}\epsilon^{ijk} \mathcal V_{ij}$).

Now we're ready for the punchline.  We now define two $3$-component vectors $\vec E$ and $\vec B$ as $q\vec E = \big(F_{01},F_{02},F_{03}\big)$ and $q\vec B = (F_{23},-F_{13},F_{12}\big)$, where $q$ is a coupling constant which quantifies the strength of the interaction.  If we do this, then we find that
$$\matrix{f^0 = \gamma \sum_{j=1}^3 F_{0j} v^j\\f^i = \gamma \big(F_{0i} - \sum_{j=1}^3F_{ij} u^j\big)} \iff \matrix{f^0 = \gamma q\vec E \cdot \vec v\\ \vec f = \gamma q(\vec E + \vec v\times \vec B)}$$
where $\times$ is the cross product in 3D.

To summarize: If we start from a relativistic viewpoint in $(3+1)$-dimensions and postulate the existence of a force $f = F\mathbf u$ which is linearly proportional to the 4-velocity $\mathbf u$, and then further split that force up into temporal and spatial components, then we are led inevitably to expressions of the form
$$\matrix{\frac{dE}{d\tau} = f^0 = \gamma q \vec E \cdot \vec v\\ \frac{d\vec p}{d\tau}= \gamma q (\vec E + \vec v \times \vec B)}$$
where $\vec E=\big(F_{01},F_{02},F_{03}\big)$ and $\vec B=\big(F_{23},-F_{13},F_{12}\big)$ are collections of components of the matrix $F_{\mu\nu}$ split into spatial and temporal parts. This splitting is frame-dependent, and so $\vec E$ and $\vec B$ mix together under a general change of frame.

Is the magnetic field really a relativistic effect of the electric field?

That's not quite right.  It's more correct to say that the electromagnetic field $F$ can be split into an electric field $\vec E$ and a magnetic field $B$, but that this splitting is frame-dependent; $\vec E$ and $\vec B$ transform like 3-vectors under rotations, but mix together if we boost to a new reference frame. It's a mistake to conclude that $\vec E$ is funadmental and $\vec B$ is not or vice-versa; rather, they are both aspects of the fundamental object $F$ looked at from a specific choice of reference frame.
A: 
Many sources have claimed that the magnetic field is a relativistic effect of the electric field

Usually the sources do not actually make this claim. It is usually something that students infer from the sources that is not actually claimed.
The magnetic field is not a relativistic effect of an electric field. A magnetic force on a single charge can always be transformed into a purely electric force in another reference frame. The same is not true of a magnetic field.
One of the invariant quantities of the electromagnetic field is $E^2-B^2$ (In units where c=1). If this quantity is negative, such as around an ordinary permanent magnet, then there is no reference frame where the magnetic field is zero. Thus the magnetic field cannot generally be seen as a relativistic consequence of the electric field.
A: 
However, those compasses were stationary, and thus are in the same frame of reference as the wire, and should not experience length contraction and not experience the repulsion/attraction by the magnetic field right?

Yes, the magnet is at rest, so it is not repelled or attracted to the wire.
Something moving inside the magnet makes trips around the magnet being either repelled or attracted to the wire at different parts of the trip, exerting different forces at different parts of the magnet, causing a torque on the magnet.
A: The simple answer is that, while the wire isn't moving, the unbound electrons within the wire, which make up the actual current, are moving, so your objection is moot. Furthermore, the wire is net neutral, so its motion relative to the compass is irrelevant, except insofar as it changes the apparent current.
