Is pressure Lorentz-invariant? Consider a system of stationary 3-D ideal gas in the rest-frame $S$. This system is described by $PV=Nk_BT.$ Also, from the principle of equipartition, $E=\frac{3}{2}Nk_BT.$
Now we introduce a boost (and a co-moving system $S'$). I am assuming $N$ and $k_B$ are Lorentz-invariant.
we get a Lorentz-Fitzgerald contraction along the axis of movement, so $V$ is downscaled by a factor of $\gamma$: $ \ \ V'=\frac{V}{\gamma}.$
As the gas is stationary in S, its momentum in S is 0 so the energy transformation gives $E'=\gamma E$ so from equipartition we get $T'=\gamma T.$
If we assume $P$ is Lorentz invariant (as I have always thought was the case) we get a contradiction! From the ideal gas law we get $T'=\frac{T}{\gamma},$ while equipartiton gives us $T'=\gamma T.$ This is obviously not the case!
We must therefore assume that pressure is non-Lorentz-invariant, 
then we get $$P'V'=Nk_BT' \implies \frac{P'V}{\gamma}=\gamma Nk_BT \implies P'=\frac{P}{\gamma^2}.\ $$
Why isn't pressure Lorentz invariant? Is it derivable from it being the trace of a stress-energy tensor, or from being force per unit area using the force transformation? Where does the factor $\frac{1}{\gamma^2}$ come from?  Were I critically wrong along the way? 
In short: How do thermodynamic sizes transform under a Lorentz boost?
EDIT: Could it be that the actual answer is that the ideal gas law as we know it (namely, $PV=Nk_BT$) is only correct for the rest frame, and it's true form for a general frame contains some factor $f(\gamma)$ such that $f(1)=1?$
 A: Pressure is part of the energy-momentum tensor. For an ideal fluid, and with the convention $g_{\mu\nu}= {\rm diag}(1,-1,-1,-1)$, this tensor can be written as
$$
 T_{\mu\nu}= (\varepsilon +p) u_\mu u_\nu -p g_{\mu\nu}.
 $$
Here $u^\mu$ is the four velocity 
$$
u^\mu=\gamma(1,{\bf v})
$$ of the fluid.
 In the local rest frame of the fluid    $u_\mu= (1,0,0,0)$, so this becomes 
$$
T_{\mu\nu}=\left[\matrix{\varepsilon  & 0& 0& 0\cr 0& p &0 & 0\cr 0&0& p & 0\cr 0&0&0&p}\right]_{\mu\nu}.
$$
So, although the  energy-momentum tensor looks quite  different in different frames, 
once  you diagonalize you will get the same $p$ and $\varepsilon$. In this sense $p$  is an invariant, because it is an eigenvalue of a matrix.
Further to answer you other point. Yes ideal gas kinetic theory only holds in the rest frame  Actually for non-deal  systems the notion of "rest frame" is not well defined. Even for a single component gas you can have a frame in which there is no energy flux (Landau frame), a frame in which there is no baryon-number  flux (Eckart frame), a frame with no entropy flux, and  frame the total momentum density is zero. In general all are different. 
