If $e^A$ and $e^B$ commute, do $A$ and $B$ commute? It is known that if two matrices $A,B \in M_n(\mathbb{C})$ commute, then $e^A$ and $e^B$ commute. Is the converse true?

If $e^A$ and $e^B$ commute, do $A$ and $B$ commute?

Edit: Addionally, what happens in $M_n(\mathbb{R})$?
Nota Bene: As a corollary of the counterexamples below, we deduce that if $A$ is not diagonal then $e^A$ may be diagonal.

I would add that if $A$ and $B$ are Hermitian matrices, $[e^A,e^B] = 0$ does imply $[A,B] = 0$. You can refer to this answer by user8675309.
To put it shortly, the eigenspaces of $e^A$ are exactly those of $A$ if $A$ is Hermitian. Since $[e^A, e^B] = 0$, $e^A$ and $e^B$ have a common eigenbasis. This basis serves as a common eigenbasis of $A$ and $B$ as well. Therefore $A$ and $B$ commute.
