Refraction and reflection in special relativity I know that special relativity postulates that the speed of light is the same no matter the inertial reference system. Consider that we have a medium with absolute refractive index $n$ ( when the medium is stationary). If we move the medium with a speed comparable to the speed of light ( so that special relativity needs to be applied) what happens to the speed of light through that medium? Is the refractive index an invariant to Lorentz transformations? Also how does a stationary observer see the speed of light through that medium? 
 A: Electromagnetic response in moving media is a tricky issue for many reasons, and an exhaustive answer is tricky. I will try to show how one could approach this problem, but there may be many more additional provisos.
Prelims:

Firstly, electric and magnetic fields are not independent, and end up being mixed when one starts switching reference frames. You have to work with the electromagnetic tensor https://en.wikipedia.org/wiki/Electromagnetic_tensor:
All latin indices are spatial ($1,2,3$ for $x,y,z$), greek indices run over temporal index as well. 
$F^{\mu\nu}=-F^{\nu\mu}$
$F^{i0} = \frac{1}{c}E^i$, where $E$ is the electric field and $c$ is the speed of light.
$F^{ij} = \epsilon^{0ijs}g_{sk}B^k$, where $\epsilon^{0ijs}$ is the Levi-Civita relative tensor, $g_{\mu\nu}=diag\left(1,-1,-1,-1\right)_{\mu\nu}$ is the metric, and $B$ is the magnetic field.
We then can introduce displacement tensor:
$D^{\mu\nu}=-D^{\nu\mu}$
$D^{i0}=cD^i$, where $D$ is the electrical displacement
$D^{ij}=\epsilon^{0ijs}g_{sk}H^k$, where $H$ is the magnetic H-field
Maxwells equations are then:
$\partial_\mu D^{\mu\nu}=0$
$\partial_\mu \left(\frac{1}{2}\epsilon^{\mu\nu\alpha\beta}g_{\alpha\alpha'}g_{\beta\beta'}F^{\alpha'\beta'}\right)=0$
Now we can introduce linear electromagnetic response for the material (for now without dispersion). Let it be that of an isotropic, non-magnetic, loss-less dielectric with refractive index $n$:
$D^i=\varepsilon_0 n^2 E^{i}$
$H^i=\mu_0^{-1} B^i$
Introduce tensor $\Xi^{\mu\nu}_{\alpha\beta}$ as:
$D^{\mu\nu}=\Xi^{\mu\nu}_{\alpha\beta}F^{\alpha\beta}$. For the material I have defined, I recon the correct expression of $\Xi$ is:
$\Xi^{\mu\nu}_{\alpha\beta}=\frac{\varepsilon_0}{2} u_\kappa u^\phi \left(\delta^{\mu\nu}_{i\phi}\delta^{i\kappa}_{\alpha\beta}n^2 + \delta^{\kappa\mu\nu}_{\phi\alpha\beta}\right)$
Where $\delta^{\dots}_{\dots}$ are the generalized Kroenecker deltas, and $u$ is the four-velocity of the medium under the consideration.

ONLY now can we consider what happens if we start boosting. Let us boost into reference frame what is moving with speed $v$ along x-direction.
The boost is given by $\Lambda^\mu_{\nu}=\left(\begin{array} & \gamma & \frac{v}{c}\gamma & 0 & 0 \\ \frac{v}{c}\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right)$
Where $\gamma=1/\sqrt{1-v^2/c^2}$
So in the new reference frame, call it $\bar{S}$, the $\Xi$ - tensor is:
$\bar{\Xi}^{\mu\nu}_{\alpha\beta}=\Lambda^{\mu}_{\mu'}\Lambda^{\nu}_{\nu'}\left(\Lambda^{-1}\right)^{\alpha'}_{\alpha}\left(\Lambda^{-1}\right)^{\beta'}_{\beta}\Xi^{\mu'\nu'}_{\alpha'\beta'}$

Results:
Lets see what the dielectric tensor is. For electric displacement we have:
$\left(\begin{array}& \bar{D}^x \\ \bar{D}^y  \\ \bar{D}^z \end{array}\right) = \varepsilon_0 \left(\begin{array}& 
n^2 & 0 & 0 \\ 
0 & 
\frac{n^2-\beta^2}{1-\beta^2} & 0 \\ 0 & 0 & \frac{n^2-\beta^2}{1-\beta^2} \end{array}\right) \left(\begin{array}& \bar{E}^x \\ \bar{E}^y  \\ \bar{E}^z \end{array}\right) + 
c\varepsilon_0 \left(\begin{array}& 0 & 0 & 0 \\ 0 & 0 & 
-\frac{\beta\left(n^2-1\right)}{1-\beta^2} 
\\ 0 & 
\frac{\beta\left(n^2-1\right)}{1-\beta^2}
 & 0 \end{array}\right) \left(\begin{array}& \bar{B}^x \\ \bar{B}^y  \\ \bar{B}^z \end{array}\right) $
where $\beta=v/c$. For magnetic H-field you have:
$\left(\begin{array}& \bar{H}^x \\ \bar{H}^y  \\ \bar{H}^z \end{array}\right) = c\varepsilon_0 \left(\begin{array}& 0 & 0 & 0 \\ 0 & 0 & 
-\frac{\beta\left(n^2-1\right)}{1-\beta^2}
\\ 0 & 
\frac{\beta\left(n^2-1\right)}{1-\beta^2}
& 0 \end{array}\right) \left(\begin{array}& \bar{E}^x \\ \bar{E}^y  \\ \bar{E}^z \end{array}\right) + 
\mu_0^{-1} \left(\begin{array}& 1 & 0 & 0 \\ 0 & 
\frac{1-n^2\beta^2}{1-\beta^2} 
& 0 \\ 0 & 0 & \frac{1-n^2\beta^2}{1-\beta^2}  \end{array}\right) \left(\begin{array}& \bar{B}^x \\ \bar{B}^y  \\ \bar{B}^z \end{array}\right) $
In such a complex material you will have not just usual plane waves, but many different kinds of waves that can propagate in a very complex fashion, so defining refractive index (just one) will not be possible unless you choose one possible type of waves.
So to answer your question. Refractive index is not a covariant quantity. Once materials start moving their electromagnetic response can change so much that you need rank 4 tensor with 36 different components to fully capture it.

EDIT. Fixed few errors. Simpler expressions. Same message
A: This is a partial and very concise answer. The refractive index is derived from the dielectric function or "constant". This "constant" in general is a tensor of which the elements may depend on frequency and direction of the EM field. To discuss its Lorentz covariant form I have to switch to Lorentz covariant notation and Einstein summation. The electromagnetic field tensor in a medium is $$F'^{\mu\nu} = \epsilon^{\mu\nu}_{\rho\sigma} F^{\rho\sigma}~.$$ Thus in general the "dielectric constant" is a very complex fourth rank Lorentz tensor with elements that are functions of frequency and field orientation.
The tensor for a moving medium is obtained by Lorentz transformation of this fourth rank tensor.
A: 
Is the refractive index an invariant to Lorentz transformations? 

No, it's not. The relativistic index of refraction is:
$$n^\prime=\frac{1+v/(nc)}{1/n+v/c}=\frac{n+\frac{v}{c}}{1+n\frac{v}{c}}$$
This refractive index can easily be calculated using the relativistic addition formula. The velocity of the light in the medium is $c/n$ and the velocity of the medium is $v$ WRT the lab observer. Therefore, the relativistic velocity addition implies:
$$w=\frac{c/n+v}{1+v/(nc)} \space ,$$
where $w$ is the speed of light WRT the lab observer. See this article. The relativistic refractive index is thus calculated to be:
$$n^\prime=\frac{c}{w}=\frac{1+v/(nc)}{1/n+v/c}=\frac{n+\frac{v}{c}}{1+n\frac{v}{c}}$$
Recall that the refractive index is the ratio of the greater velocity ($c$) to the smaller one ($w$). 
However, remember that this index of refraction is for when the light travels parallel to the motion direction of the medium and perpendicular to the interface. Calculations are slightly complicated for other directions.
