Lorentz invariance of Maxwell's equations in matter I know that Maxwell's equations of electromagnetism are Lorentz invariant in a vacuum. But what about in a generalized medium, e.g. a metal, a rubber, a dielectric, a magnet? I have read it comes down to whether the electric and magnetic polarizations, $M$ and $P$, are themselves Lorentz invariant. (Note: I am ignoring gravity.) My feeling is that they must be, albeit some researchers might use approximations that are not. So can anyone answer my question:
Are Maxwell's equations in a medium Lorentz invariant?
 A: 
Are Maxwell's equations in a medium Lorentz invariant?

No, they're not. An electromagnetic wave can have a speed $v<c$ in a medium. That means you can choose a frame in which the wave's velocity is zero. A zero-velocity wave is a solution in that frame, but not in the rest frame of the medium. Therefore the equations are not form-invariant under a Lorentz boost.
A: The aspect of Maxwell's equations that fails to be "fully covariant" in the presence of matter are the constitutive relations.  Specifically, if you write out Maxwell's equations in the presence of matter,
\begin{align*}
\nabla \cdot \vec{D} &= \rho_f & \nabla \times \vec{E} &= - \frac{\partial \vec{B}}{\partial t} \\
\nabla \cdot \vec{B} &= 0 & \nabla \times \vec{H} &= \vec{J}_f + \frac{\partial \vec{D}}{\partial t}
\end{align*}
you need to supplement these with a set of relationships between the auxiliary fields $\vec{D}$ & $\vec{H}$ and the "real" fields $\vec{E}$ and $\vec{B}$.  For example, we commonly assume that $\vec{D} = \epsilon \vec{E}$ and $\vec{H} = \frac{1}{\mu} \vec{B}$;  from this, we can show that electromagnetic waves will travel through the material at the same speed $v = c/\sqrt{\epsilon \mu}$ in all directions; the fields all obey the wave equation with a characteristic speed $v$.
But these simple-looking relationships ($\vec{D} \propto \vec{E}$, $\vec{H}\propto \vec{B}$) are a frame-dependent statement, and they will not necessarily hold in another reference frame in which the medium is moving.  As an example of this, let's assume that the new "primed" reference frame is moving in the positive $x$-direction with respect to our original frame.  The fields in the two reference frames are related to each other by (for example)
\begin{align*}
E_y &= \gamma (E'_y + v B'_z)  & D'_y &= \gamma(D_y - v H_z) &
B_z &= \gamma (B'_z + v E'_y) 
\end{align*}
(note that the transformation among $\vec{D}$ and $\vec{H}$ is the same as the transformation between $\vec{E}$ and $\vec{B}$.) We then have
\begin{align*}
D'_y &= \gamma(D_y - vH_z) \\
&= \gamma \left(\epsilon E_y - v \frac{1}{\mu} B_z \right) \\
&= \gamma \left[\epsilon \gamma (E'_y + v B'_z) - v \frac{1}{\mu} \gamma (B'_z + v E'_y) \right] \\
&= \gamma^2 \left(\epsilon - \frac{v^2}{\mu}\right) E'_y + \gamma v \left( \epsilon - \frac{1}{\mu} \right) B'_z \neq \epsilon E'_y. 
\end{align*}
Thus, even if $\vec{D}$ is proportional to $\vec{E}$ in the rest frame of the material, this does not imply that $\vec{D}'$ will be proportional to $\vec{E}'$ in a non-rest frame.1  Instead, the constitutive relations are more complicated in this frame;  $\vec{D}$ will depend on both $\vec{E}$ and $\vec{B}$, as will $\vec{H}$.  
We could in principle still use Maxwell's equations and these new constitutive relations to write down a second-order differential equation for $\vec{E}$ or $\vec{B}$ alone.  But what we'll find is a wave-like equation in which the speed of wave propagation differs in different directions.  And if we take a valid propagation velocity vector in the rest frame and transform it into our new frame, it'll line up exactly with a valid propagation velocity of light according to our new wave-like equation.

1 Unless $\epsilon = 1/\mu$, but then we're talking about a medium in which all waves travel at the speed of light anyhow.
A: Maxwell is not Lorentz invariant in matter because  the material selects a preferred reference frame — the one in which the lump of matter is at rest.  
Of course you can make it all look Lorentz covariant by including the local four velocity $u^\mu$  of the piece of matter in the equations  for the dielectric constant and the magnetic permeability, and by  defining 
$$
 E_\mu = F_{\mu\nu}u^{\nu}, \quad B_\mu = \frac 12 \epsilon_{\mu\nu\sigma\tau} u^\nu F^{\sigma\tau}.   
 $$ 
to be the ${\bf E}$ and ${\bf B}$ fields in the frame moving with the matter — but that extra $u^\mu$ makes everything rather complicated.  It's best to avoid all this unless you really want to to do relativistic fluid/continuum  mechanics such as investigating the magnetic field on a neutron star or the accretion disc of a black hole.
A: Maxwell's equations in vacuum are Lorentz covariant, not invariant. In a moving medium they are also covariant, but as you stated, mostly they are not written in a covariant manner. Any physical system is Lorentz covariant else special relativity would fail.
A: I'm going to pick on your choice of word.
The optical axis or axes of a crystal are only Lorentz invarant if the direction of the boost is parallel, antiparallel, or perpendicular to the optical axis/axes.  (In fact, birefringence was the first phenomenon to come to mind when I read your question.)
A: Electromagnetics in a material is an example of spontaneous symmetry breaking.
Consider a crystal in a rest frame. The lattice structure breaks the rotation symmetry, and indeed, crystals are often birefringent. Similarly, a ferromagnet has a preferred direction determined by the bulk magnetization.
The underlying governing equations for the crystal and the ferromagnet are Maxwell's equations and a many-particle Schrödinger equation for nuclei and electrons, and they certainly have rotation symmetry, but the crystalline or ferromagnetic ground state does not, and the nearby (in energy) excitations do not either. The ground state has spontaneously broken the symmetry.
