Does the energy of a photon change in different inertial systems? Given a photon with energy $E$ is some system S what would be it's energy in a system S' with a constant velocity v relative to S?
I have no idea where to start from. I think you need to calculate the wavelength and than apply some form of a doppler affect however i dont know any version of doppler affect that works in SR
 A: A photon with energy $E$ propagating in the $+x$ direction in (for simplicity) 1+1D spacetime has four-momentum (well... two-momentum) $$p^\mu=\begin{bmatrix}E\\E\end{bmatrix}$$ where $c = 1.$ You can get this from 1) that $p^0 = E$ (the time component of any four-momentum is energy) 2) that $p^\mu p_\mu=0$ (the four-momentum is lightlike because photons are massless), and that 3) the spatial part of the four-momentum is in the direction of propagation.
Taking that $p^\mu$ to have been measured in $S$, the Lorentz transform from $S$ to a frame $S'$ moving at velocity $v$ (in the $+x$ direction) is given by $${\Lambda^\mu}_\nu=\begin{bmatrix}\gamma&-\gamma v\\-\gamma v&\gamma\end{bmatrix}\qquad(\gamma=\frac{1}{\sqrt{1-v^2}}).$$
The transformation is then $p^\mu\to p'^\mu={\Lambda^\mu}_\nu p^\nu,$ (note that this is matrix multiplication) and in particular we're only actually interested in $$E'=p'^0={\Lambda^0}_\mu p^\mu=\gamma(1-v)E=\sqrt{\frac{1-v}{v+1}}E.$$
While this derivation of the transformation rule for photon energy is not as edifying as the one in @Buzz's linked Wikipedia page, I think it is valuable to also know the "mechanical" way of answering all such questions about Lorentz transforms. You express the quantities you are interested in in terms of Lorentz covariant objects (scalars, like proper time, or four-vectors, like four-momentum, or four-tensors, like the electromagnetic field, etc.), and then you just apply Lorentz transforms to them (via matrix multiplication). That way you don't have to think as hard to come up with a more specific derivation.
Also, do note that the equation for the Doppler effect (with velocities measured along the direction of propagation), $$f_r=\frac{c-v_r}{c-v_s}f_s,$$ works pretty much as-is in SR. The only nuance is that the quantities in it all "belong" to one reference frame, so you have to add appropriate factors of $\gamma$ to get the quantities you want. Specifically, if an observer at rest ($v_s=0$) in frame $S$ observes a frequency of $f_s$, then by the ordinary equation for the Doppler effect an observer moving at relative velocity $v=v_r$ receives frequency $f_r=(1-v)f_s,$ and this is correct as long as you realize $f_r$ is still measured in $S$. Because time in the receiver's rest frame $S'$ runs slower by a factor of $\gamma$, the frequency as measured in $S'$ is $f_r'=\gamma f_r,$ which gives the same formula as before.
