$A$ and $B$ are square matrices of the same size.
$A$ is symmetric, and $B$ is not symmetric.
How can I prove (or disprove) that there is no matrix C such that
$C^T = C^{-1}$ and $A = C^{-1} B C$?
I tried manipulating $A = C^{-1} B C$ using $C^T = C^{-1}$ and the fact that $A$ is symmetric to get to the conclusion that $B = B^T$ (thus proving that the existence of $C$ would contradict the initial assumptions), but I didn't manage to get that.
From $A=C^{-1}BC$ you get, multiplying by $C$ on the left and by $C^{-1}$ on the right, $B=CAC^{-1}$. Since $A$ is symmetric, $$ B^T=(CAC^{-1})^T=(C^{-1})^TA^TC^T=C^{TT}AC^{-1}=CAC^{-1}=B. $$