Showing: if G acts on A by conjugation then the stabilizer of A in G is the Normalizer of A in G. This is a theorem from Dummit & Foote text-

The number of conjugates of a subset $ A$ in a group $G$ is the index of the normalizer of $A$,$\vert G:N_G(A) \vert$.

The highlighted text is a part of the proof of the above theorem-
Let $a \in A$ be, then stabilizer of $A$ in $G$ is $G_a =${$g \in G:g.a=a$} 
$\implies$ $G_a =${$g \in G:gag^{-1}=a$}
Normalizer of $A$ in $G$, $N_G(A)=$ {$g \in G:gAg^{-1}=A$}.
I'm getting some sort of confusion in notations due to which i'm unable to link $G_a$ and $N(A)$. Hence, not getting how to reach to the desired result. 
Need some hint!
thank you!
 A: Let $G$ acting on a set $X$ by $(g,a)\mapsto g.a$, for each $a\in
X$ the stabilizer of $a$  under  ( or relatively to) this action
is the maximal sub-group $N$ of $G$,  s.t $g.a=a, \forall g\in H$;
the standard  notation  of this stabilizer is $St(a)$.
When $G$ acts on the set of all its subgroup by
conjugation the stabilizer $St(H)$ of a subgroup $H$ relatively
to this action is exactly the normalizer  $N(H)$ of $H$ in $G$
this result by the definition.
Now for $St(H)=\{g\in G \mid gHg^{-1}=H\}$ this is a maximal subgroup of
$G$ in which $H$ is normal so is $N(H)$.
If $G$ act on himself by conjugation, then $St(a)$ is denoted by $C_a$
and if we look at the definition, a priori there is no link between $St(H)$ and $St(a)$ for $a\in H$, also as it is well known that  $H_1\cup H_2$
is not necessarily a subgroup. But $\cap_{a\in H} C_{a}$ is a subgroup of $G$ that is the centralizer of $H$ in $G$ and denoted by $Z_G(H)$, in particular $Z_G(H)\cap H=Z_H(H)=Z(H)$ the centralizer of $H$.
Now $[G:N(H)]$ is the cardinal of right coset $(G/N(H)_g$ and this set is mapped bijectively to the  orbit $O_H=\{gHg^{-1},g\in G\} $ this orbit is exactly the set of conjugates of $H$ in $G$.
A: For the proof of the theorem, consider the action in the setting
$$G\times\wp(G)\to\wp(G),$$
where $\wp(G)$ is the $G$'s power-set,
defined, of course, as $g\cdot A=gAg^{-1}$, then, by the bijection ${\rm Orb}\leftrightarrow{\rm G/St}$, we have
$\#{\rm Orb}(A)=[G:N_G(A)]$, since
$$N_G(A)=\{g\in G\mid gAg^{-1}=A\}={\rm St}(A).$$
