Given that $\pmb{AB} \ne\pmb{BA}$ and $\pmb{BA}^{-1} \ne\pmb A^{-1}\pmb B$, are there conditions where $\pmb{ABA}^{-1}=\pmb B$ ?
I am working in the context of a continuous, LTI system $\pmb{\dot x}(t)=\pmb{Ax}(t)+\pmb{Bu}(t)$ where the eigenvalues of $\pmb A$ have strictly negative real parts. I know that in at least one case $\pmb Ae^{\pmb At}\pmb A^{-1}=e^{\pmb At}$. An inverse question could be are there cases where $\pmb Ae^{\pmb At}\pmb A^{-1}\ne e^{\pmb At}$ ?
No. If $ABA^{-1}=B$, then right multiplying by A gets us $$ABA^{-1}A=BA$$ $$ABI=BA$$ $$AB=BA$$
Similarly, if $A$ is invertible, then you can start with $AB=BA$ and right multiply by $A^{-1}$ to reverse the steps. So if $A$ is invertible, the two conditions are equivalent.