"The p-power bounding" right multiplication in a finite group G under some special conditions

Question

[Roughly speaking, the following question considers a special setting in which we want to prove a property in the form of $ord(g \sigma)\ |\ p^k$.]

The Problem in Detail:
Let $G$ be a finite group, and $H \leq G$ any subgroup with order $|H|=p^k$ (where $p$ is a prime, and $k \in \mathbb{N}$. Note that $H$ may not be a Sylow $p$-subgroup). Let $g \notin H$ be an element in $G$ s.t. $ord(g)\ |\ p^k $; and $\sigma\in G$ be s.t. for every integer $n$ we have $g^n \sigma g^{-n} \in H$. We claim that $ord(g \sigma)\ |\ p^k$.
(Edit: The previous claim is $ord(g) = ord(g \sigma)$, which is wrong due to the counterexample provided by @jpvee.)

My Attempts:
I can virtually do nothing other than expanding $(g \sigma)^l$, which does not contribute much.
I am quite confused on how to even start out the first step properly.

Can anyone please give me some hints on how I should proceed?
Thank you very much for your help! :)

Answer 1

I have found a reference for proving this claim. My claim is in fact a weaker version of the first lemma given in the article "On A Theorem of Frobenius" by Richard Brauer in 1969.

Since the proof is clearly presented in the original article (which can be easily accessed). I would not copy it down here :)

Answer 2

Sorry to disappoint you, but your conjecture is false. One counterexample I could come up with is the following:

Let $G:=S_4$ be the symmetric group on four points, let $H:=\langle\,(1,2,3,4),(1,3)\,\rangle$ be a Sylow $2$-subgroup. Now choose $g:=(1,4)$ and $\sigma:=(1,2)(3,4)$.

Then $g\not\in H$, and $g^n\sigma g^{-n}\in H$ for all $n\in\mathbb{Z}$ (since $\sigma$ is an element of the Klein subgroup which is a normal subgroup of $G$ contained in $H$).

On the other hand, we have $g\sigma=(1,3,4,2)$ and thus $ord(g)=2\neq4=ord(g\sigma)$