Is the following true?
Let $A \in \mathbb{R}^{n\times n}$ and suppose that $v(t)$ be a (continuous) eigenvector of $\text{Exp}(At)$ for all $t$. Then $v(t)$ is equal to a constant vector $w$ if and only if $w$ is an eigenvector of $A$.
Of course, if $w$ is an eigenvector of $A$ then it is trivially an eigenvector of $\text{Exp}(At)$ as well.
What about the other direction?
Suppose $v(t)=w$. Since $v(t)$ is an eigenvector we have that
$$e^{At}v(t) = \lambda(t) v(t)$$
and take derivatives on both sides
$$Ae^{At}v(t) + e^{At}v'(t) = \lambda(t) v'(t) + \lambda'(t) v(t)$$
Now plug in $v'(t)=0$ then $v(t)=w$:
$$A\left(e^{At}v(t)\right) = \lambda'(t)v(t)$$
$$A(\lambda(t)v(t)) = \lambda'(t)v(t)$$
$$Aw = \frac{\lambda'(t)}{\lambda(t)}w$$
which is an eigenvalue equation for $w$ with $A$. One can go even farther by taking one more derivative and showing that
$$0 = \left(\frac{\lambda'(t)}{\lambda(t)}\right)'w \implies \frac{\lambda'(t)}{\lambda(t)} = k$$
for arbitrary $w\neq 0$. Thus $\lambda(t) = C \cdot e^{kt}$.
$\textbf{EDIT}$: By Severin's suggestion below, one might wonder if it is always well defined to divide by $\lambda(t)$. The determinant of $e^{At}$ is the product of its eigenvalues, but we have that
$$\det e^{At} = e^{t\cdot\text{tr} A} \neq 0$$
so $0$ cannot be an eigenvalue of $e^{At}$.