Why 4-momentum is not transforming as vector when time is reversed? $p^{\nu} = (E/c,\textbf{p})^T$
After time reversal Lorenz transformation ( $t'=-t,r'=r$) it becomes:
$p'^{\nu} = (E/c,\textbf{-p})^T$
But if we multiply transformation matrix by old coordinates, we obtain the opposite result:
$(E'/c,\textbf{p}')^T =diag(-1,1,1,1)*(E/c,\textbf{p})^T = (-E/c,\textbf{p})^T $
So, 4-momentum is not a vector...
 A: It's because the time reversal operator $T$ is anti-unitary. You need to understand exactly how the transformation of the momentum is derived. The unitary operator that generates translations $U(a)$ satisfies
$$
T^{-1} U(a)T = U ( {\cal T}a) , \qquad {\cal T} = diag(-1,+1,+1,+1). \tag{1}
$$
Now, we recall that $U(a) = \exp ( i a^\mu P_\mu )$ where $P_\mu$ is the translation generator. Expanding equation (1) to first order in $a^\mu$, we find
$$
T^{-1} ( i P^\mu )T = {\cal T}^\mu{}_\nu ( i  P^\nu )
$$
But, now we recall that $T$ is anti-unitary so $T i T^{-1} = - i$. Using this, we find
$$
T^{-1} P^\mu T  = - {\cal T}^\mu{}_\nu P^\nu 
$$
The rest of the argument is as follows. Let $| p \rangle$ be a momentum eigenstate
$$
P^\mu | p \rangle = p^\mu | p \rangle
$$
Then, consider the state $T | p \rangle$. This satisfies
$$
P^\mu ( T | p \rangle ) = T T^{-1} P^\mu T | p \rangle = T ( - {\cal T}^\mu{}_\nu P^\nu  | p \rangle = - {\cal T}^\mu{}_\nu p^\nu ( T  | p \rangle ) 
$$
It follows that the state $T | p \rangle$ has momentum $- {\cal T}^\mu{}_\nu p^\nu$. This is often written concisely by stating that under time reversals
$$
p^\mu \to - {\cal T}^\mu{}_\nu p^\nu \quad \implies \quad ( E , \vec{p} ) \to ( E , - \vec{p} ). 
$$
A: This is only a proof for massive particles as:
$$p^\nu = m\frac{\text{d}x^\nu}{\text{d}\tau}$$
where $\tau$ is the proper time, $m$ the mass and $x^\nu$ the position of the particle in spacetime.
Under the transformation $t'=-t$ and $\vec{r}\,'=\vec r$ the proper time is unchanged! Thus, you only get a change in sign of the $\nu=0$ component. Therefore, your second expression holds true:
$$(E'/c,\textbf{p}')^T =diag(-1,1,1,1)*(E/c,\textbf{p})^T = (-E/c,\textbf{p})^T$$
The error in your first expression is you use the derivative with respect to coordinate time not the proper time for your definition of the 4-momentum.
