What are sufficient conditions for a matrix to have the same eigenvectors as its exponential? If $\boldsymbol{A}$ is a square matrix, then it is straightforward to show that each eigenvector to $\boldsymbol{A}$ is also an eigenvector to $e^\boldsymbol{A}$.
On the other hand, an eigenvector to $e^\boldsymbol{A}$ is not necessarily an eigenvector to $\boldsymbol{A}$. For example, the eigenvectors to
$$
\boldsymbol{A}=
\begin{bmatrix}
2\pi i& 0 \\
0 & 4\pi i
\end{bmatrix}
$$
are the scalar multiples of $\begin{bmatrix}1 & 0\end{bmatrix}^T$ and $\begin{bmatrix}0 & 1\end{bmatrix}^T$, while any non-zero vector is an eigenvector to
$$
e^\boldsymbol{A}=
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}.
$$
Are there any interesting conditions which guarantee that $\boldsymbol{A}$ and $e^\boldsymbol{A}$ have exactly the same eigenvectors?
 A: I think sufficient conditions are that $A$ and $e^A$ have the same the number of distinct eigenvalues. 
What we really need to show is that the eigenspaces of $A$ and $e^A$ are the same:
Let $\sigma(A)=\{\lambda_1,\dots,\lambda_m\}$ denotes the spectrum of $A$ (the set of eigenvalues of $A$) and let $V_k$ be the eigenspace of $A$ associated with $\lambda_k$ (that is, the set of vectors $v$ such that $Av=\lambda_k v$). From the definition of $e^A$ it follows that 
$$\{e^{\lambda_1},\dots,e^{\lambda_m}\}\subseteq\sigma(e^A).$$
Furthermore, if $U_k$ denotes the eigenspace of $e^A$ associated with $e^{\lambda_k}$, it is easy to see that $V_k\subseteq U_k$. Since, $A$ and $e^A$ have the same number of eigenvalues, the $e^{\lambda_k}$s are distinct. Which implies that if $v$ belongs to any $V_i$ with $i\neq k$, then $v$ does not belong to $U_k$. Since $\mathbb{R}^n$ is spanned by the union of the $V_k$s, it is also spanned by the union of the $U_k$s. Combining the last two phrases we have that $U_k=V_k$ which gives the desired result.
Edit: If you can think of a simple way to argue that $\{e^{\lambda_1},\dots,e^{\lambda_m}\}\supseteq\sigma(e^A)$ (I can't at the moment), then it is easy to argue that the above are necessary conditions as well: If they don't have the same number of eigenvalues, then $e^{\lambda_i}=e^{\lambda_j}$ for some $i,j$. In which case you can pick $v\in V_i$ and $w\in V_j$ and you have that $v+w$ is an eigenvector of $e^A$ but not of $A$.
A: You can assume without loss of generality that $A$ is in Jordan normal form, so
$$A = \begin{bmatrix} J_1 & 0 & 0 & \dotsb & 0\\
0 & J_2 & 0 & \dotsb & 0\\
\vdots & & \ddots & \ddots & \vdots\\
0 & \dotsb & 0 & J_{r-1} & 0\\
0 & \dotsb & \dotsb & 0 & J_r\end{bmatrix}$$
with Jordan blocks $J_\rho$ to the eigenvalue $\lambda_\rho$. Then you have
$$e^A = \begin{bmatrix} e^{J_1} & 0 & 0 & \dotsb & 0\\
0 & e^{J_2} & 0 & \dotsb & 0\\
\vdots & & \ddots & \ddots & \vdots\\
0 & \dotsb & 0 & e^{J_{r-1}} & 0\\
0 & \dotsb & \dotsb & 0 & e^{J_r}\end{bmatrix}$$
and each $e^{J_\rho}$ has the form $e^{\lambda_\rho}\cdot I_{k\times k} + N$ with a strictly upper (or lower, depends on what you consider the Jordan normal form) triangular $k\times k$ matrix with all $1$s on the off-diagonal, so the Jordan normal form of $e^A$ has the same structure as that of $A$, only the eigenvalues $\lambda_\rho$ are replaced with the eigenvalues $e^{\lambda_\rho}$.
The eigenspaces of $e^{A}$ are thus exactly the eigenspaces of $A$, unless $A$ has eigenvalues $\lambda_1 \neq \lambda_2$ with $e^{\lambda_1} = e^{\lambda_2}$, in which case you have
$$E_A(\lambda_1) \oplus E_A(\lambda_2) \subset E_{e^A}(e^{\lambda_1}),$$
with equality if there is no $\lambda_3 \notin \{\lambda_1,\,\lambda_2\}$ with $e^{\lambda_3} = e^{\lambda_1}$. Generally, you have
$$E_{e^A}(e^{\lambda_1}) = \bigoplus_{e^\lambda = e^{\lambda_1}} E_A(\lambda).$$
So the answer is that $A$ and $e^A$ have exactly the same eigenspaces if and only if $A$ has no pair of distinct eigenvalues $(\lambda_1,\lambda_2)$ with $e^{\lambda_1} = e^{\lambda_2}$, or equivalently, if and only if no two eigenvalues of $A$ differ by an integral multiple of $2\pi i$.
