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


This is what I have done so far, but I am stuck and not sure if I am in the right direction:

$xabx^{-1} = ba$

Multiply both sides on the right by $x$.

$xabx^{-1}x = bax$.

Now the $x^{-1}x$ cancel out: $xab = bax$. 

This is where I am stuck because I am not sure how to solve for an $x$ in $G$.
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.)
If I put $~X = b~$ or $~X = a^{-1} ~$(consider as self inverse)than what we get

I get that the term     $~xabx^{-1} = ba ~$, become   $~ ba = ba~$

Which gives that it is satisfied.

As my opinion