If the matrix exponential is unitary, is the exponent necessarily skew-Hermitian?

Question

If $A$ is skew-Hermitian, then $e^A$ is unitary. But is the converse true? That is, if $e^A$ is unitary, is $A$ necessarily skew-Hermitian?

Answer 1

This is just a proof of @user1551's comment:

The eigenvalues of $A=\begin{pmatrix}0&-\pi\\4\pi&0\end{pmatrix}$ are $\pm2\pi i$, so $$S^{-1}AS=\begin{pmatrix}2\pi i&0\\0&-2\pi i\end{pmatrix},$$ and $$e^A=e^{SAS^{-1}}=Se^AS^{-1}=S\begin{pmatrix}2^{2\pi i}&0\\0&e^{-2\pi i}\end{pmatrix}S^{-1}=SS^{-1}=I.$$ However, $A$ is not skew-Hermitian.