$C'XC=X\Rightarrow CXC'=X?$

Question

This question is derived from the lorentz transformation.

if $X=diag(1,1,1,-1)$ and $C$ is an invertible matrix, $C'XC=X$. can we conclude that $CXC'=X$ ?

Answer 1

Yes. This is because $X^2=I$. Suppose that $AXB=X$ for some square matrices $A$ and $B$ (whether $A=B'$ is irrelevant). Then $B$ must be nonsingular. Therefore \begin{aligned} AXB=X &\Rightarrow AX=XB^{-1}\\ &\Rightarrow X(AX)X=X(XB^{-1})X\\ &\Rightarrow XA=B^{-1}X\\ &\Rightarrow BXA=X. \end{aligned}