If in a group $G$, we have $a^5 = e$ and $aba^{-1} = b^2$ for some $a$, $b$ in $G$, then what is the order of $b$?

It is not hard to see that $a^nba^{-n}=b^{2^n}$. Thus, $b^{31}=1$. Note that $31$ is prime and so $o(b)=31$ or $b=1$.

Here is a more general result.

If $G$ is a group and $a$,$b$ $\in$ $G$ with $bab^{-1} = a^r$ for some $r$ $\in$ $\mathbb{N}$, then $b^jab^{-j} = a^{r^j}$ for all $j$ $\in$ $\mathbb{N}$

Proof : We use induction. The base case is trivial. Let the hypothesis be true for some natural number $k$. We have :

$b^{j+1}ab^{-(j+1)} = b(b^jab^{-j})b^{-1} = ba^{r^j}b^{-1} = (bab^{-1})^{r^j} = (a^r)^{r^j} = a^{r.r^j} = a^{r^{j+1}}$

QED.

Now you can proceed as Babak did. This is taken from an exercise in Hungerford's Algebra.