I shall modify an argument for energy-mass equivalence due to Fritz Rohrlich to be valid inside a dielectric medium with refractive index $n$ and light velocity $c_n=c/n$.

Let us assume that inside the dielectric photons obey the following dispersion relationship:

$$

\begin{eqnarray*}

\omega &=& c_nk,\\

\mathbf{p} &=& \hbar\ \mathbf{k},\\

E &=& \hbar\ \omega.

\end{eqnarray*}

$$

Therefore inside the medium the magnitude of the photon momentum, $p$, is related to its energy, $E$, by the relation:

$$p=\frac{E}{c_n}.$$

Now imagine a body with mass $M$ at rest.

Suppose it simultaneously emits two pulses of light, one to the left and one to the right, each with energy $E/2$.

As the momenta of the light pulses are equal and opposite then the body remains at rest.

Now consider the experiment from the perspective of an inertial frame moving with velocity $-\mathbf{v}$ to the left.

Initially the body has momentum $\mathbf{P}=M\mathbf{v}$ to the right.

After the photons are emitted the right-going photon has its frequency/energy Doppler blueshifted by the factor $(1+v/c_n)$ and the left-going photon has its frequency/energy redshifted by the factor $(1-v/c_n)$.

The momentum of the right-going photon is changed by

$$\Delta \mathbf{P} = \frac{\mathbf{v}}{c_n}\frac{E}{2c_n}.$$

The momentum of the left-going photon is also changed by $\Delta \mathbf{P}$.

Therefore the momentum of the body after the emission of the photons is

$$\mathbf{P'}=M\mathbf{v}-2\Delta\mathbf{P} =\left(M-\frac{E}{c_n^2}\right)\mathbf{v}.$$

Therefore the change in the body's mass is equal to the total photon energy lost divided by $c_n^2$ (rather than $c^2$).