Supose $N$ is a normal subgroup of order $p^n$. Then show that $\sigma (N)=N$

Question

Let $G$ be a group of order $p^nm, \ p \nmid m$. Supose $N$ is a normal subgroup of order $p^n$. Then show that $\sigma (N)=N$ for any automorphism $\sigma \in \text{Aut}(G)$.

I have tried in the way-

Suppose $H$ be the smallest subgroup of $G$ containing both $\sigma(N)$ and $N$.

My target is to show $|H|=p^n$.

One way to show this, we need to show $H\subset N$.

Suppose $H \nsubseteq N$, then there exists at least an element, say, $g \in G \setminus \{H\}$ such that order of $g$ doesn't divide $p^n$.

How to show contradiction here ?

Edit:

I got the following answers but if we can prove the question in my direction of approach.

Answer 1

Ok, here is a proof/hint which avoids using the Sylow theorems (again, not sure why this is desirable). We have that $$\lvert \sigma(N) N \rvert = \frac{\lvert \sigma(N) \rvert \lvert N \rvert}{\lvert \sigma(N) \cap N \rvert} = \frac{p^{2n}}{\lvert \sigma(N) \cap N \rvert}.$$

Since $N$ is normal, $\sigma(N)N$ is a subgroup of $G$, so its order divides $p^n m$. What does this tell you about the order of $\sigma(N) \cap N$?