If $(e^{\alpha(t)})'=\alpha'(t)e^{\alpha(t)}$ then $\alpha(t), \alpha'(t)$ commute? Let $\alpha(t)$ be a smooth path of real $n \times n$ matrices. (Formally $\alpha:(-\epsilon,\epsilon) \to M_n(\mathbb{R})$).

If $(e^{\alpha(t)})'=\alpha'(t)e^{\alpha(t)}$ for every $t$ or $(e^{\alpha(t)})'=e^{\alpha(t)}\alpha'(t)$ for every $t$, is it true that $\alpha'(t), \alpha(t)$ commute?

What happens if we only know that $(e^{\alpha(t)})'|_{t=0}=\alpha'(0)e^{\alpha(0)}$ or $(e^{\alpha(t)})'=e^{\alpha(0)}\alpha'(0)$? Does this imply that $\alpha'(0), \alpha(0)$ commute?
I don't even know what happens in the case where $\alpha(t)=V+tW$ for some fixed $V,W \in M_n$. In fact, I don't know whether or not $e^{V+tW}=e^Ve^{tW}$ for every $t$ implies that $V,W$ commute.
I do know that $e^{Xt}e^{Yt} = e^{(X+Y)t} \implies [X,Y] = 0$ (just differentiate the equality twice at $t=0$).
Edit:
By using the identity for the derivative of the exponential, we have that
$$ (e^{\alpha(t)})'=e^{\alpha(t)}\frac{1-e^{-\text{ad}_{\alpha(t)}}}{\text{ad}_{\alpha(t)}} (\alpha'(t)),$$  so
$$(e^{\alpha(t)})'=e^{\alpha(t)}\alpha'(t) \iff \frac{1-e^{-\text{ad}_{\alpha(t)}}}{\text{ad}_{\alpha(t)}} (\alpha'(t))=\alpha'(t).$$
I am not sure this really helps.
 A: Here's a counterexample. Take
$$\alpha(t)=\gamma\begin{pmatrix}-1&t\\0&1\end{pmatrix}\in\mathbb C^{2\times 2}$$
where $\gamma$ is a non-zero solution to $\gamma e^\gamma=\sinh\gamma.$ To be concrete take $\gamma=\tfrac12(W_1(-e^{-1})+1)$; here $W_1$ is a particular non-principal branch of Lambert's W function. According to Wolfram Alpha $\gamma\approx -1.04 + 3.73 i.$
$\alpha(t)$ can be thought of as living in $M_4(\mathbb R)$ by considering $\mathbb C^2$ as a real vector space.
By direct computation it does not commute with $\alpha'(t).$ I will show that $(e^{\alpha(t)})'=\alpha'(t)e^{\alpha(t)}.$
Since $\alpha(t)^2=\gamma^2I$ where $I$ is the identity matrix, and the power series for $\cosh(z)$ and $z^{-1}\sinh(z)$ only use even powers of $z,$
\begin{align}
e^{\alpha(t)}
&=\cosh(\alpha(t))+\sinh(\alpha(t))\\
&=\cosh(\alpha(t))+\alpha(t)\alpha(t)^{-1}\sinh(\alpha(t))\\
&=\cosh(\gamma)I+\alpha(t)\gamma^{-1}\sinh(\gamma)\\
&=\cosh(\gamma)I+\begin{pmatrix}-1&t\\0&1\end{pmatrix}\sinh(\gamma).\\
\end{align}
Hence
\begin{align}
\alpha(t)'e^{\alpha(t)}
&=\gamma\begin{pmatrix}0&1\\0&0\end{pmatrix}\Bigl(\cosh(\gamma)I+\begin{pmatrix}-1&t\\0&1\end{pmatrix}\sinh(\gamma)\Bigr)\\
&=\gamma\begin{pmatrix}0&1\\0&0\end{pmatrix}\cosh(\gamma)+\gamma\begin{pmatrix}0&1\\0&0\end{pmatrix}\sinh(\gamma)\\
&=\gamma\begin{pmatrix}0&1\\0&0\end{pmatrix}e^\gamma\\
&=(e^{\alpha(t)})'
\end{align}
as required.
