Show that if $(\sigma^r)^i$ is an $n$-cycle, then $\sigma^r$ is an $n$-cycle

Question

I'm trying to show that if the power of $\sigma$ is an $n$-cycle, then $\sigma$ itself is an $n$-cycle. This seems to be something that is privy to a proof by contradiction but I can't seem to see how.

Answer 1

Let $\sigma = (123)(45)$. Then $\sigma^2 = (132)$ is a $3$-cycle but $\sigma$ is not a $3$-cycle.

Answer 2

The original version stated that $\sigma$ is a cycle and $\sigma^i$ an $n$-cycle. The way I read it is that the task was to prove that the lengths of those two cycles must match. If we don't know that $\sigma$ is a cycle (the question was edited to read this way), then the claim is false. See Zoe's answer for a counterexample.

---

(Hints/answer to the original version)

If $\sigma$ is an $m$-cycle, then the $m$ objects moved by $\sigma$ are not mapped to themselves by any power $\sigma^i, 0<i<m$. Therefore $\sigma^i$ cannot be a shorter cycle (because it moves all those $m$ objects). However, it can be a product of several disjoint shorter cycles.