A nice counterexample has already been presented.

More generally, for any $W$ of the form $e^{X}$, where $X$ is an element of a Lie algebra, it follows from an identity related to the BCH formula that

$e^X A e^{-X} = \sum_{n=0}^{\infty}\frac{[X^n, A]}{n!}$.

So in such cases (as in the examples) the statement will clearly fail if $A$ and $X$ do not commute, and their commutator is also Hermitian.

A nice counterexample has already been presented.

More generally, for any $W$ of the form $e^{X}$, where $X$ is an element of a Lie algebra, it follows from an identity related to the BCH formula that

$e^X A e^{-X} = \sum_{n=0}^{\infty}\frac{[(X)^n, A]}{n!}$.

So in such cases (as in the examples) the statement will clearly fail if $A$ and $X$ do not commute, and their commutator is also Hermitian.