Can the equality $e^{-tY}Me^{tY} = e^{tX}M $ be shown by showing it only to 1st order? (Lie representations)

Question

We have that A and B belong to different representations of the same Lie group. The representations have the same dimension. X and Y are elements of the respective Lie algebra representations. $$A = e^{tX}$$ $$B = e^{tY}$$ We want to show, for a specific matrix M $$B^{-1} M B = AM$$ Does it suffice to show this to first order? $$\left(1 -tY + \dots \right)M \left(1 + tY + \dots \right) = \left(1 + tX + \dots \right)M$$ In other words is $$-YM + MY = XM$$ sufficient to show $$B^{-1} M B = AM$$ for all t?

Answer 1

I see no reason whatsoever why one would expect such a thing to be true. It would means that $B^{-1}MB=AM$would imply (by substituting $nt$ for $t$) $B^{-n}MB^n=A^nM$, which seems totally wrong.

Let us try an example. Take $B=(\begin{smallmatrix}a^{-1}&0\\0&a\end{smallmatrix})$ and $M=(\begin{smallmatrix}1&1\\0&1\end{smallmatrix})$, then $B^{-1}MB=(\begin{smallmatrix}1&a^2\\0&1\end{smallmatrix})$ which we can write as $AM$ for $A=(\begin{smallmatrix}1&a^2-1\\0&1\end{smallmatrix})$. Now $B^{-2}MB^2=(\begin{smallmatrix}1&a^4\\0&1\end{smallmatrix})\neq(\begin{smallmatrix}1&2a^2-1\\0&1\end{smallmatrix})=A^2M$. It fails.

To answer to the comment, you can do the same infinitesimally. Take $Y=(\begin{smallmatrix}-1&0\\0&1\end{smallmatrix})$ and still $M=(\begin{smallmatrix}1&1\\0&1\end{smallmatrix})$, then $-YM+MY=(\begin{smallmatrix}0&2\\0&0\end{smallmatrix})=XM$ for $X=(\begin{smallmatrix}0&2\\0&0\end{smallmatrix})$. With $B=e^{tY}=(\begin{smallmatrix}e^{-t}&0\\0&e^t\end{smallmatrix})$ and $A=e^{tX}=(\begin{smallmatrix}1&2t\\0&1\end{smallmatrix})$ this gives $B^{-1}MB=(\begin{smallmatrix}1&e^{2t}\\0&1\end{smallmatrix})\neq AM$.