Prove that if for all $g \in G$ that $gNg^{-1} \subseteq N$, then we must have for all $g\in G$ that $gNg^{-1}=N$ I feel kind of shaky about this proof so I'd really appreciate some input. I know this is equivalent to some other forms (which I'll prove afterward so please don't spoil it). Here's the statement again:

Prove that if for all $g \in G$ that $gNg^{-1} \subseteq N$, then we must have for all $g\in G$ that $gNg^{-1}=N$.

Attempt: Assume $gNg^{-1}\subsetneq N$. Then $\exists gng^{-1}, gn'g^{-1} \in gNg^{-1}$ with $n\neq n'$ such that
$$gng^{-1}=gn'g^{-1}$$
but that implies $n=n'$, thus no such elements exist and $gNg^{-1}=N$.
Is this correct? I know I should've instead tried to show if $n \in N$ then $n\in gNg^{-1}$ but this seemed to accomplish the same.

The statement is true for all groups, not just finite ones.  What you've written here is not well explained, it looks like you're showing that conjugation is injective and so it's a bijection.  But this style proof will only work when $N$ is finite.
Instead, what happens if you conjugate $N$ by $g^{-1}$ instead of $g$?
