Let $a$ and $b$ belong to a group $G.$ Find an $x$ in $G$ such that $xabx^{-1}=ba.$

Try letting $x = a^{-1}$, so that $x^{-1} = (a^{-1})^{-1} = a$.

Or, if you prefer, let $x = b$, so $x^{-1} = b^{-1}$.

(Since $a, b \in G$, so are $a^{-1}, b^{-1} \in G$, since $G$ is a group and a group is closed under taking inverses.)