For matrix-valued functions, does $\frac d {dt} \exp(A(t)) = A'(t)\exp(A(t))$ imply $A'(t)A(t)=A(t)A'(t)$? The converse (that $A$ and $A'$ commuting implies $(\exp(A)' = A'\exp(A)$) is easy to show from the series for $\exp$. In a class I'm TA for, one question on the students' exam was to find an example of a non-constant matrix function $A(t)$ such that $\exp(A)$ solves $\dot \Phi = \dot A \Phi$. All the (correct) solutions I've seen satisfy $A\dot A = \dot A A$, but I am wondering if this is actually necessary.
 A: Lie theory permits to show that the derivative of the exponential map is given by
$$
\frac{\mathrm{d}}{\mathrm{d}t}\exp(A) = \exp(A)\psi(-\mathrm{ad}_A)\dot{A}, 
   \verb+ +\mathrm{with}\verb+ + 
   \psi(x) = \frac{e^x-1}{x} 
   \verb+ +\mathrm{and}\verb+ +
   \mathrm{ad}_A = [A,\,\cdot\,],
$$
where $[X,Y] = XY-YX$ is the anticommutator. The only way to enforce $\frac{\mathrm{d}}{\mathrm{d}t}\exp(A) = \dot{A}\exp(A)$ is to have $\mathrm{ad}_A(\dot{A}) = [A,\dot{A}] = 0$. Indeed, one has${}^1$
$$
\dot{A}\exp(A) = \exp(A)\exp(-A)\dot{A}\exp(A) = \exp(A)\exp(-\mathrm{ad}_A)\dot{A}
$$
and, by comparison with the expression for $\frac{\mathrm{d}}{\mathrm{d}t}\exp(A)$ above, $\exp(-\mathrm{ad}_A)\dot{A} = \psi(-\mathrm{ad}_A)\dot{A}$, whence
$$
\dot{A} = \exp(\mathrm{ad}_A)\psi(-\mathrm{ad}_A)\dot{A} = \frac{\exp(\mathrm{ad}_A)-1}{\mathrm{ad}_A}\dot{A} = \sum_{n=0}^\infty\frac{\mathrm{ad}_A^n}{(n+1)!}\dot{A},
$$
which implies the condition $\displaystyle\sum_{n=\color{red}{1}}^\infty\frac{\mathrm{ad}_A^n\dot{A}}{(n+1)!} = 0$. Of course, this condition is easily satisfied when $\mathrm{ad}_A\dot{A} = [A,\dot{A}] = 0$ (i.e. when $A$ commutes with its derivative), but there are other occurrences, such as the one brought to light in DarkMalthorp's answer.

${}^1$N.B. : this step is nothing else than the well-known Baker-Campbell-Hausdorff formula, which states that $e^BAe^{-B} = A + [B,A] + \frac{1}{2!}[B,[B,A]] + \frac{1}{3!}[B,[B,[B,A]]] + \ldots$ or, more succinctly, $\mathrm{Ad}_{e^B} = e^{\mathrm{ad}_B}$ within Lie theory.
A: The answer is no:
Let $c$ be nonzero complex number such that $$
\frac{e^c-1}c = 1
$$
The possible values of $c$ are given by $-1-W(-1/e)$ for all the non-real branches of the Lambert $W$ function. For example, we can take $$
c=-1-W_{1}(-1/e) \approx 2.08884 - 7.46149i
$$
Then take $$
A(t) = \begin{bmatrix}0&0\\t&c\end{bmatrix}
$$
Then $$
\exp(A(t)) = \begin{bmatrix}1&0\\ \frac{e^c-1}c t & e^c\end{bmatrix} = \begin{bmatrix}1&0\\ t & e^c\end{bmatrix} 
$$
Observe $$
\frac d{dt} \exp(A(t)) = \begin{bmatrix}0&0\\1&0\end{bmatrix} =  \begin{bmatrix}0&0\\1&0\end{bmatrix}\begin{bmatrix}1&0\\ t & e^c\end{bmatrix} = A'(t) \exp(A(t))
$$
and $$
[A,\dot A] = \begin{bmatrix}0&0\\c&0\end{bmatrix} \ne 0
$$
Similarly, we could build a real example by $$
A(t) = \begin{bmatrix}0&0&0&0\\0&0&0&0\\t&0&2.08884&-7.46149\\0&t&7.46149&2.08884\end{bmatrix}
$$
I wonder whether a 2x2 or 3x3 real example is possible. I will update this answer if I find one.
