Is X Always Commutative with Y in Matrix Invariance?

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Since many schools have just started the school year, I'd like to begin September's POTW with an important, yet simple problem.

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Suppose $X$ and $Y$ are $n \times n$ complex matrices such that $X$ is $Y$-invariant, i.e., $e^{zY}Xe^{-zY} = X$ for all $z\in \Bbb C$. Prove that $X$ and $Y$ commute.
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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

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No one solved this week's problem correctly. You can read my solution below.

Spoiler

Differentiate the equation $e^{zY}Xe^{-zY} = X$ with respect to $z$ to obtain

$$Ye^{zY}Xe^{-zY} - e^{zY}XYe^{-zY} = 0,$$

then set $z = 0$ to get $YX - XY = 0$, or $YX = XY$.