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.
 A: I just want to point out a general(maybe not the most general) way of constructing infinitely many counterexamples in $M_2(\mathbb{R})$. It is not hard to prove for a $2\times2$ traceless matrix $X$, we have
$$e^X=\cos(\sqrt{\det X})I+\frac{\sin\sqrt{\det X}}{\sqrt{\det X}}X,$$
where $\frac{\sin 0}{0}$ is to be understood as 1 for the second term. Thus to have $e^X=I$ we just need some traceless $X$ having $\det X=(2n\pi)^2,n\in\mathbb{Z/\{0\}}$, and this is a fairly easy job. Most of the matrices you construct in this way won't commute with each other. All the previously given $M_2(\mathbb{R})$ counterexamples fall into this category.
A: No. Let $$A=\begin{pmatrix}2\pi i&0\\0&0\end{pmatrix}$$ and note that $e^A=I$. Let $B$ be any matrix that does not commute with $A$.
A: Here's an example over $\mathbb{R}$, modeled after Harald's answer: let
$$A=\pmatrix{0&-2\pi\\ 2\pi&0}.$$
Again, $e^A=I$. Now choose any $B$ that doesn't commute with $A$.
A: Another example: 
$$A=\pmatrix{0&-2\pi\\ 2\pi&0}, \textrm{      }B=\pmatrix{0&-2\pi\\49\cdot 2\pi&0}$$
This is a counterexample of the following statement:
$$e^{A+B}=e^Ae^B\Longrightarrow AB=BA$$
since 
$$e^A=e^B=e^{A+B}=I$$ but 
$$AB=4\pi^2 \pmatrix{-49&0\\ 0&-1},$$
$$BA=4\pi^2 \pmatrix{-1&0\\0&-49}.$$
A: 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.
