Given $X, Y \in \mathbb{R}^{n \times n}$ and a $C^1$ surjective diffeomorphism $F: \mathbb{R}^n \to \mathbb{R}^n$, suppose that $e^{tX} \circ F = F \circ e^{tY}$ for all real $t \in [0, \infty)$. I wish to verify that there exists an invertible matrix $M$ such that $MA = BM$.
To begin, I noticed that $F(0) = 0$ need not hold. So, let $a \in \mathbb{R}^n$ such that $F(0) = a$. However, $G(x) = F(x) - a$ should also satisfy the same conditions that $F$ does. It is clear that $G$ is $C^1$ diffeomorphism from $\mathbb{R}^n$ onto $\mathbb{R}^n$. Unfortunately, it is not clear to me that $e^{tX} \circ G = G \circ e^{tY}$. I have tried \begin{align*} G \circ e^{tY}(x) &= (F - a) \circ e^{tY}(x) \\ &= F \circ e^{tY}(x) - a \\ &= e^{tX} \circ F(x) - a \\ &= e^{tX} \circ G(x) & (\text{This is the questionable step.}) \end{align*} By taking $t = 0$ at the end, this would then allow us to obtain \begin{align*} D(G \circ e^{tY}) De^{tY} = D(e^{tX} \circ G) DG &\implies D(G \circ e^{tY})(0) De^{tY}(0) = D(e^{tX} \circ G)(0) DG(0) \\ &\implies DG(0) e^{tY} = e^{tX} DG(0) \\ &\implies \frac{d}{dt} \left(D(G)(0) e^{tY}\right) = \frac{d}{dt} \left(e^{tX} DG(0)\right) \\ &\implies DG(0) Y e^{tY} = Xe^{tX} DG(0) \\ &\implies DG(0) Y = X DG(0). \end{align*} The invertibility of $DF$ then allows us to conclude that $DG(0)$ is invertible.
I thought about this a little more closely. The equality holds; the steps are shown below: \begin{align*} G \circ e^{tY}(x) &= (F - a) \circ e^{tY}(x) \\ &= F \circ e^{tY}(x) - a \\ &= F \circ e^{tY}(x) - F(0) \\ &= F \circ e^{tY}(x) - F \circ e^{tY}(0) \\ &= e^{tX} \circ F(x) - e^{tX} \circ F(0) \\ &= e^{tX} \circ (F(x) - a) \\ &= e^{tX} \circ G(x) \\ \end{align*}