Find the order of an element of finite group Let $G$ be a finite group and $g,h\in G-\{1\}$ such that $g^{-1}hg=h^2$.
In addition $o(g)=5$ and $o(h)$ is an odd integer. Find $o(h)$.
I know from a previous exercise that if there exists a natural number $i$ such that $g^{-1}hg=h^i$ then for all $n\in \mathbb{N}$, $g^{-n}hg^n=h^{i^n}$.
I thought I could use this fact somehow, but so far no luck.
Please give me a hint.

Square both sides
$$
g^{-1}h^2 g = h^4
$$
Now replace $h^2$
$$
g^{-1}(g^{-1} h g) g  = g^{-2} h g^2 = h^4
$$
Again square and expand
$$
g^{-3}hg^3 = g^{-2}h^2g^2 = h^8.
$$
By repeating this process we find
$$
g^{-k} h g^k = h^{2^k}
$$
for $k>0$
so in particular
$$
h = g^{-5}hg^5 = h^{32}.
$$
Hence $h^{31} = 1$. Thus $O(h)$ divides $31$. But $31$ is prime, so $O(h) = 1$ or $O(h) = 31$. Since we assumed $h \neq 1$, we find $O(h) = 31$.
Note here that the given information $O(h)$ is an odd integer was an unnecessary assumption. We could immediately see this by noting that the order of $h$ and any conjugate, e.g. $g^{-1}h g = h^2$, are the same. Thus $h,h^2$ have the same order, so $2$ is not a divisor of $O(h)$. 
