Equivalent definition of normalizer of a group. Need help in proof. The definition of normalizer of a non-empty set $A$ in group $G$ is given by $$N_G(A) = \{g\in G\:|\:gAg^{-1} = A\}$$

This definition is from Abstract Algebra by Dummit and Foote.

Another definition I have seen around the internet (here and in this video) is $$N_G(A)=\{g\in G\:|\:gag^{-1}\in A\:\forall\: a\in A\}$$

I try to prove these equivalent. So we need $$gAg^{-1}=A \iff gag^{-1}\in A \:\forall\: a \in A$$

Here is my partial proof:

Assume $gAg^{-1}=A$. Then $gag^{-1}\in gAg^{-1}=A \:\forall \:a \in A$.

Conversely, assume $gag^{-1}\in A \:\forall\: a \in A$. Let $x\in gAg^{-1}$. Then, $\exists\:a_x \in A$ such that $x=ga_xg^{-1}\in A$. It follows that $gAg^{-1}\subseteq A$.

Now , let $y \in A$. This implies $gyg^{-1} \in A$. (This is where I'm stuck). 

If I could somehow prove $y \in A \implies g^{-1}yg\in A$, then the result would from the hypothesis. Please let me know if I'm headed in the wrong direction.

I have seen this MathSE post. The definition of the normalizer is a slightly modified version of the first one stated here. However, I would like to know how I can proceed from where I'm stuck.
$gAg^{-1}$ = {$gag^{-1} : a\in A$}

Therefore, $gAg^{-1} = A$ occurs iff $g^{-1}Ag=A$ which both imply $\forall a\in A\; gag^{-1},g^{-1}ag\in A$

Now, assume $\forall a\in A\; gag^{-1}\in A$. From this we have $gAg^{-1}\subset A$. Choose $b\in A$ and note that $y=g^{-1}bg\in A$ and therefore $$b=gyg^{-1}\in gAg^{-1}$$

implying $A\subset gAg^{-1}$. Therefore, $gAg^{-1} = A$ iff $\forall a\in A\; gag^{-1}\in A$
$gAg^{-1}$ = {$gag^{-1} : a\in A$}

Therefore, $gAg^{-1} = A$ occurs iff $g^{-1}Ag=A$ which both imply $\forall a\in A\; gag^{-1},g^{-1}ag\in A$

Now, assume $\forall a\in A\; gag^{-1}\in A$. From this we have $gAg^{-1}\subset A$. Choose $b\in A$ and note that $y=g^{-1}bg\in A$ and therefore $$b=gyg^{-1}\in gAg^{-1}$$

implying $A\subset gAg^{-1}$. Therefore, $gAg^{-1} = A$ iff $\forall a\in A\; gag^{-1}\in A$