Invariance under conjugation, equivalent in Lie Group and Lie Algebra? Is the following true? 
$ e^X Y e^{-X} = Y \Leftrightarrow [X,Y]=0$ .
From right to left you can show it with a corollary from the Baker–Campbell–Hausdorff formula. But in the other direction? 
I could just give some shaky physicists argument like "it's the infinitesimal version so it should hold" however i'm unsure about it's validity.
 A: It's true that $[X,Y]=0 \implies e^X Y e^{-X} = Y$, but not conversely. To prove this implication, assume $[X,Y]=0$, and use the power series for the matrix exponential:
\begin{align*}
e^{X}Y &= \left(\sum_{n=0}^\infty \frac{1}{n!}X^n\right)Y= 
\left(\sum_{n=0}^\infty \frac{1}{n!}X^nY\right)=
\left(\sum_{n=0}^\infty \frac{1}{n!}YX^n\right)\\
&=
Y\left(\sum_{n=0}^\infty \frac{1}{n!}X^n\right)=Ye^X.
\end{align*}
Here's a counterexample that shows $e^X Y e^{-X} = Y$ does not imply $[X,Y]=0$:
$$ X = \left(\begin{matrix}0&\pi\\-\pi& 0 \end{matrix}\right),\quad
Y = \left(\begin{matrix}0&1\\1& 0 \end{matrix}\right).$$
Here $X$ and $Y$ do not commute, but $e^X$ is minus the identity, which does
commute with $Y$.
If you assume the stronger hypothesis 
that $e^{tX} Y e^{-tX} = Y$ for all $t\in \mathbb R$ (this is what @Robin Ekman 
is implicitly assuming in another answer here),
it then follows that $[X,Y]=0$. A simple way to prove this is to differentiate the identity $e^{tX}Ye^{-tX}\equiv Y$ with respect to $t$ and set $t=0$.
A: Suppose that $e^{tX} Y e^{-tX} = Y$. The left-hand side is the definition of $\operatorname{Ad}_{e^{tX}} Y$. We have $$
0 = \frac{d}{dt}\bigg|_{t=0}  \operatorname{Ad}_{e^{tX}} Y = \operatorname{ad}(X)Y = [X, Y]
$$
where the first equality is the hypothesis, the second is the definition of $\operatorname{ad}$ and the last a known theorem. You can find a proof for example in Manifolds and Differential Geometry by Lee, chapter 5. The consequences of this theorem as worked out by Lee on the following pages are along the lines of $\operatorname{ad}$ being the infinitesimal version of conjugation and the exponential map connects them.  
