Suppose every eigenvalue $\lambda$ of $X$ satisfies $|\lambda|<2\pi$. If $\exp(X)v=v$ for some $v\in \mathbb C^n$ then $Xv=0$. Let $X\in M_n(\mathbb C)$. Suppose every eigenvalue $\lambda$ of $X$ satisfies $|\lambda|<2\pi$. If $e^Xv=v$ for some $v\in \mathbb C^n$ then $Xv=0$.
I know that if I suppose $e^Xv=v$, then 1 is an eigenvalue of $e^X$, so then $\log e^X=X$ has eigenvalue $\log 1=0$. So $Xv=0$. However, I know there are some requirements to be able to use the matrix logarithm, and that I have not used all my hypothesis, so this is suspect, but I am not sure what else to do.
I have the hint to use the series of $\frac{e^x-1}{x}$ but I'm not sure how to fit this in either.
 A: Here is a slightly unsatisfactory answer.
Let $f(x) = e^x-1$. Note that $f^{(k)}(x) = e^x$ for $k \ge 1$.
Suppose $J_{\lambda,m}$ is a Jordan block of size $m$ with $\lambda$ on the diagonal.
Then (see, for example, https://en.wikipedia.org/wiki/Jordan_matrix#Functions_of_matrices),
$e^{J_{\lambda,m}}-I = \begin{bmatrix} e^\lambda-1 & {1 \over 2!} e^\lambda  &
{1 \over 3!}e^\lambda & \cdots & {1 \over (m-1)!} e^\lambda \\
0 & e^\lambda-1 & {1 \over 2!} e^\lambda & \cdots  & {1 \over (m-2)!} e^\lambda \\
\vdots & \vdots & \ddots & & \vdots\\
0 & 0 & 0 & \cdots & e^\lambda-1
\end{bmatrix}$.
From this we see that if $v \neq 0$ and $(e^{J_{\lambda,m}} -I)v = 0$ then $e^\lambda =1$
and $v=e_k$.
Since $e^\lambda =1 $, we must have $\lambda \in 2 \pi i \mathbb{Z}$ and so, by assumption, $\lambda = 0$. Hence $J_{\lambda,m} v = 0$. Hence if $(e^{J_{\lambda,m}} -I)v = 0$ then  $J_{\lambda,m} v = 0$.
Let $J$ be a Jordan normal form of $X$. Then $e^J = \operatorname{diag}(e^{J_{\lambda_1,m_1}},...,e^{J_{\lambda_k,m_k}})$ and from the above it follows
that if $(e^J-I)v=0$ then $Jv = 0$.
A: Suppose $Av = v$, we have
$$e^{2\pi iA}v = e^{2\pi i}v = v$$
Let $X = 2 \pi inA, (n \in \mathbb Z)$, then
$$
e^{X}v = e^{2\pi i n A}v = v
$$
Which fit the requirements but
$$
Xv = 2\pi inAv = 2\pi i n v
$$
$X$ also have an eigenvalue of $\lambda = 2\pi in$ where $\lvert\lambda\rvert = 2\pi n$
From this example we can see that:
$$
e^Xv = \lambda v \Rightarrow Xv = (log \lambda + 2 \pi i n)v  (n \in \mathbb Z)
$$
This is why we need $\lvert\lambda\rvert < 2\pi$
A: Let $X=D+N$ where $D$ is diagonalizable, $N$ is nilpotent and $ND=DN$. Then $e^X=e^De^N=e^D(I+N+\cdots)=e^D(I+M)$ where $M$ is nilpotent, $\ker(M)=\ker(N)$ and $MD=DM$.
$e^Xv=v$ $\implies$ $e^Dv=v$, and $e^DMv=0$, that implies $Nv=0$ and finally $Xv=Dv$. 
We may sassume $D=diag(\lambda_i)$ where $|\lambda_i|<2\pi$; then 
$e^Dv=v$ IFF for every $i$, $e^{\lambda_i}v_i=v_i$ IFF for every $i$, $v_i=0$ or $e^{\lambda_i}=1$, that is $\lambda_i=0$.
Thus, for every $i$, $\lambda_iv_i=0$; therefore $Dv=0$ and we are done.
