Is the speed of light in all media independent of reference frame? We know that, the speed of light in vacuum is independent of reference frame. I read the reason for this fact from Why and how is the speed of light in vacuum constant, i.e., independent of reference frame?. Similarly, is the speed of light in a medium of refractive index $n \neq 1$ also independent of reference frame?
I tried to frame my question visually. So, let us consider the following image:
 
The entire setup is in a medium of refractive index not equal to unity. The light source is at rest with respect to the medium. The violet coloured squares below the light beam numbered $1,2,$ and $3$ are speed detectors which we can use to determine the speed of light beam with respect to the reference frame attached to them. $1$ is at rest with respect to the light source. $2$ moves parallel to the light beam and towards the source with a speed of $v$ and $3$ moves away with the same speed. Now my question is, is the speed of light beam detected by the three detectors equal? If it's vacuum we know they are equal. But what happens in this case?
Image Courtesy: My Own Work :)
 A: The formula for adding velocities is $\frac {v_1 + v_2}{1+\frac{v_1v_2}{c^2}}$. Plugging $c$ in for $v_1$ yields $c$ for any $v_2$. Plugging in a value of $v_1$ other than $c$ and a nonzero value for $v_2$ results in a value other than $v_1$.
In other words, anything observed to be travelling at $c$ in one reference frame will be observed to be travelling at $c$ in all reference frames. Anything observed to be travelling at a velocity other than $c$ in one reference frame will be observed to be travelling at other velocities (and by "other", I mean different from its observed velocity in the first reference frame, not just different from $c$) in other reference frames. 
In particular, anything travelling at a velocity other than $c$ has a rest reference frame in which it will be observed to have zero velocity. If Detector 3 in your diagram is travelling at $\frac c n$, then the light wave will appear to be a standing wave.
A: Let's use the Lorentz transformation to calculate this. Let S be a reference frame, in which a coordinate system $(t,x)$ is used. Let S$'$ be a frame moving in the $x$ direction relative to S at speed $v$. The Lorentz transformation asserts that if event A is at $(t,x)$ then its coordinates in S$'$ are given by
\begin{eqnarray}
t' &=& \gamma(t - v x / c^2) \\
x' &=& \gamma(x - v t)
\end{eqnarray}
where $\gamma = (1-v^2/c^2)^{-1/2}$.
Now consider a pulse of light that starts from $(0,0)$ and propagates at the speed $c/n$ relative to S. For example, there could be some water at rest in S, and $n$ is the refractive index of this water. At time $t$ in S, such a pulse will have reached the event $(t,x) = (t, ct/n)$. The coordinates of this event relative to S$'$ are given by
\begin{eqnarray}
t' &=& \gamma(t - v (ct/n) / c^2) \\
x' &=& \gamma((ct/n) - v t)
\end{eqnarray}
Now the coordinates of the starting point are $(0,0)$ in both frames, so we can find the speed of this pulse relative to S$'$ by using the distance traveled divided by the time elapsed:
$$
\mbox{speed relative to S}' = \frac{x' - 0}{t' - 0} = \frac{c/n - v}{1 - v (c/n) / c^2}
 = \frac{c/n - v}{1 - v / n c}.
$$
If you are familiar with the formula for addition of velocities, you could find this same result by applying it. We now find that when $n=1$ the speed relative to S$'$ is equal to $c$, but when $n \ne 1$ the speed relative to S$'$ is not equal to $c/n$. 
The above formula gives the speed that will be measured by your detector number 3, if we take it that the detector works in the usual way by measuring distances and times in its own rest frame. The result for detector number 2 will be
$$
 \frac{c/n + v}{1 + v / n c}.
$$
Another question that arises is the speed of the light relative to the water. That is just $c/n$. As soon as one says "relative to the water" then to calculate it you must use the rest frame of the water. End of story. But someone might ask instead, what is the rate of change of the distance between the light pulse and something floating in the water? If, relative to some given frame, the water is flowing at speed $w$ and the light is moving at speed $u$, then the answer to this question is $u-w$. 
A: The media itself constitutes a preferred reference frame in which the speed of light is $v=c/n$. The speed of light in a reference frame that moves relative to the media will change anisotropically.
A: The speed of light only differs when comparing it in two mediums with a different refraction index. With the added changes we can say that the detector with $v=v+$ will detect the light as going $-v$ slower than $c$ and vice versa.
A: The other answers give some good results how to arrive at the correct result from the reference frame of the water: We just relativistically add the two speeds to see that our speed of light measured changes with the detector movement.
However, relativity means the same result (up to time dilation, obviously) should come out when we look at it from the detector's reference frame. So what does that look like?
We have our light pulse, starting at some point away from the detector, then moving towards it until it reaches it and is registered (and its speed measured somehow). While doing so, it moves through a refractive medium (the water) that itself moves relative to the reference frame.
Therefore, the water (from the perspective of the detector) undergoes Lorentz contraction, meaning it becomes "shorter" in the direction of movement. For a flow of water, this means it becomes denser: The molecules along the direction of flow appear squashed and tighter together.
While the exact processes of how the interaction of light in matter results in a refractive index are complicated, it suffices for our explanation that it's dependent on density: If there's more matter in a volume to interact with, the light will be slowed down more. Thus, what changes from the perspective of the detector is $n$: Moving water has a different refractive index compared to water at rest.
