If $XYZ=ZXY$ does $e^Xe^Ye^Z=e^Ze^Xe^Y$? It is well known that if $X,Y$ are commuting matrices, then their exponential commute:
$$XY=YX\quad\implies\quad e^Xe^Y=e^Ye^X.$$
Now, I am wondering if the following generalization holds:

Question: If $XYZ=ZXY$, does $e^Xe^Ye^Z=e^Ze^Xe^Y$?

Note that if $Z$ commutes with both $X$ and $Y$, then it is obvious.
 A: OP is asking if

$$ [XY,Z]~=~0\qquad  \stackrel{?}{\Rightarrow}\qquad [e^Xe^Y,e^Z]~=~0~? \tag{1}$$

In a comment above, user1551 has already pointed out obvious counterexamples if $X=0$ xor $Y=0$. 
Here we will give a counterexample with invertible $2\times 2$ matrices, namely the Pauli matrices:
$$ X~=~ i\pi \sigma_x, \qquad  Y~=~ \frac{i\pi}{2} \sigma_y, \qquad Z~=~ \frac{i\pi}{2} \sigma_z, \tag{2}$$
$$ e^X~=~ -{\bf 1}_{2\times 2}, \qquad  e^Y~=~  i\sigma_y, \qquad e^Z~=~ i\sigma_z. \tag{3}$$
Now $XY$ is proportional to $\sigma_z$ and therefore commutes with $Z$; while $e^Xe^Y$ is proportional to $\sigma_y$, and hence anticommutes with $e^Z$ (rather than commutes).
A: If $X$ and $Y$ commute, $$e^Y e^X = e^X e^Y = e^{X+Y}$$
So $Z$ needs to commute with $X+Y$ for that to be true in this particular case, and not with $e^{XY}$.
An easy counterexample is $Y=0$, $X$ and $Z$ such that $[X,Z]\neq 0$. The condition is still fulfilled but, obviously, $e^X e^Z \neq e^Z e^X$ in the general case.
A: If $(XY)Z=Z(XY)$ we can only say that 
$$
e^Ze^{XY}=e^{XY}e^Z=e^{Z+XY}
$$
