How to find a non-singular matrix $T$ such that $ST\Lambda T^{-1}$ is Hermitian

Question

Let $S$ be Hermitian and $\Lambda$ be an arbitrary complex matrix of the same size as $S$. Is it possible to find an invertible matrix $T$ such that $ST\Lambda T^{-1}$ Hermitian? Does $T$ have any explicit formula?

Answer 1

The answer is NO. Up to an orthonormal change of basis, we may assume that $S=diag(s_j)$ where $s_j\in\mathbb{R}$. Let $L=T\Lambda T^{-1}$ and $spectrum(\Lambda)=(l_j)$.

Assume that $S$ is invertible. Then $SL=L^*S^*=L^*S$ and $L^*=SLS^{-1}$. Since $L,L^*$ are similar, necessarily, the sets $(l_j)_j$ and $(\overline{l_j})_j$ are equal.

Then a counter-example is obtained with $S=diag(1,2),\Lambda=diag(0,i)$ where $i^2=-1$.