$e^{iA}e^{iB}=e^{iB}e^{inA} \Longrightarrow e^{itA}e^{itB}=e^{itB}e^{intA}$?

Question

Let $A$ and $B$ be selfadjoint operators ($A$=$A^\ast$, $B$=$B^\ast$) on a Hilbert space and $n\in\mathbb{N}$ such that $e^{iA}e^{iB}=e^{iB}e^{inA}$, where $i=\sqrt{-1}$.

Does then $e^{itA}e^{itB}=e^{itB}e^{intA}$ hold true for any real $t$?