Let $G$ be a finite group and let $g, h, x \in G$. Does $g^{k} x h^{k} = x$ for all $k \in \mathbb{Z}$ imply that $g$ and $h$ are inverse and that they commute with $x$?
If $g = h^{-1}$ and $gx = xg$ then $g^{k} x h^{k} = x$ will certainly always hold but I'm unsure why the converse should or shouldn't hold and I can't think of a counter example.
It's enough to say $gxh=x$, since that implies $g^kxh^k=x$ for all other $k$.
And no, it doesn't imply $g=h^{-1}$, or that either of them commute with $x$. For instance, in $S_3$, we have $$ (123)(12)(123)=(12) $$ and $(123)\neq(123)^{-1}$, and $(123)(12)\neq(12)(123)$.
However, if we also know that $g=h^{-1}$, then that implies that $g$ and $x$ commute. And if we know that $g$ and $x$ commute, then that implies that $g=h^{-1}$.