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.
 A: Let $g \in G$. We know that $gNg^{-1} \subseteq N$. We want to show that $$gNg^{-1} \supseteq N.$$
Let's just multiply this containment on the left by $g^{-1}$ and on the right by $g$ to get that we want to show $$N \supseteq g^{-1}Ng.$$
So, we just need to show that $g^{-1}Ng \subseteq N$. But we already know that for all $x \in G$ that $xNx^{-1} \subseteq N$. Hence let $x = \ldots$? (Once you know what to pick for $x$, you will have finished a proof.)

The way your attempt goes, I would instead say this: Assume $gNg^{-1} \subsetneq N$.  Thus there exists $n_0 \in N$ such that for all $n \in N$, $gng^{-1} \neq n_0$. But this can't actually be true because all we need to do is consider $n = g^{-1}n_0g$.

As can be seen in the comments below, I interpreted the 'for all' quantifier a little differently than what the OP originally wrote. Indeed, the statement as it was written does not hold for all infinite groups. Here is a counterexample: 
Let $G = \mathbb{Z}[1/2] \rtimes \mathbb{Z}$. Let $(1,0)$ be a normal generator of $\mathbb{Z}[1/2]$, and let $(0,1)$ generate the $\mathbb{Z}$ on the right side of the semidirect product. Let $N = \langle (1,0) \rangle$. Notice that $N$ is not a normal subgroup of $G$. Indeed,  if we conjugate $N$ by $(0,1)$ we get $\langle (2,0) \rangle$, which is a proper subset of $N$. But if we conjugate $N$ by $(0,-1)$, then we get $\langle (1/2, 0) \rangle$, which is not a subset of $N$.
A: 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 $g$?
