- Let $S$ be a nonzero complex symmetric unitary matrix ($S\neq 0$, $S = S^T$, $S^HS=Id$)
- Let $B$ be a nonzero skew-hermitian matrix ($B\neq 0$, $B^H = -B$)
Can their product $P\triangleq SB$ be complex skew-symmetric ($P^T = -P$)? (I would say no)
Using the fact that $S = VV^T$ for a unitary $V$ and $B = UDU^H$ for a unitary $U$ and a purely imaginary diagonal matrix $D$, I could show that $P$ can be complex symmetric if $V = U^*$, but I could not show that it cannot be skew-symmetric.
Since $B$ is skew-Hermitian, it can be written as $K+iH$ for some real skew-symmetric matrix $K$ and real symmetric matrix $H$. The condition that $(SB)^T=-SB$ is therefore equivalent to $KS=SK$ and $HS=-SH$. It should be easy to construct a feasible solution $(K,H,S)$. Consider e.g. $$ K=\pmatrix{0&0\\ 0&0\\ &&0&-1\\ &&1&0},\ H=\pmatrix{1&0\\ 0&-1\\ &&0&0\\ &&0&0},\ S=\pmatrix{0&1\\ 1&0\\ &&1&0\\ &&0&1}. $$