Let $\|\cdot\|$ be the spectral norm, i.e., largest singular value. The condition number of an invertible complex matrix $X$ is defined as $\kappa(X):=\|X\|\|X^{-1}\|$.

I am able to prove

**Proposition** Let $A, B$ be $n\times n$ positive definite matrices. If $X$ is an $n\times n$ invertible matrix such that $AXB$ is Hermitian, then \begin{eqnarray*}
\|X^{-1}AXB\|\le \kappa(X) \|AB\|.
\end{eqnarray*}

In particular, if moreover $X$ in the above proposition is unitary, then $\|X^{-1}AXB\|\le \|AB\|$.

I wonder if $\kappa(X)$ in the above proposition can always be replaced with $1$.