Is this identity regarding matrix exponential correct?

Question

Given that A is a constant matrix:

$e^{-At}Be^{At} = e^{-At}e^{At}B = B$

Answer 1

Consider $A=\mathrm{diag}(1,2).$ Then $e^{At}=\mathrm{diag}(e^{t},e^{2t}),$ so $$e^{-At}Be^{At}=\begin{bmatrix}b_{1,1}&b_{1,2}e^{t}\\ b_{2,1}e^{-t}&b_{2,2}\end{bmatrix}\neq B,$$ whenever $t\neq 0.$

Answer 2

It's true if $A$ and $B$ commute, but not otherwise (for general $t$).