Q:

Defining a set of automorphisms of a group.

I'm reading an article that defines the next set:
$N=\{ x\in N_S(P)| c_g\circ c_x \circ {c^{-1}_g} \in \mbox{Aut}_S(gPg^{-1}) \}$
Where $S$ is a subgroup of a bigger finite group $G$, $gPg^{-1}$ is a subgroup of $S$ in which $P$ is a subgroup of $S$ too (the $g\in G$ is the same as the $g$ mentioned in the set), $c_n$ is the conjugate function with the element $n$, $N_S(P)$ is the normalizer of $P$ under $S$ and Aut$_S(gPg^{-1})$ are only the automorphisms that are conjugations with elements in $S$. I was trying to understand the set, and I made this:
If $x\in N$, exists a $s\in S$ such that for $p_1\in P$, exists $p_2\in P$ such that:
$c_g\circ c_x\circ c^{-1}_g(gp_1g^{-1})= gxg^{-1}(gp_1g^{-1})gx^{-1}g^{-1}=s(gp_2g^{-1})s^{-1}=gp_3g^{-1}$
Using the two right equalities we have that, as $c_s$ is an automorphism, is clear that is necessarily true that
$$ s(gp_2g^{-1})s^{-1}=gp_3g^{-1} $$
and it means that $g^{-1}s(gp_2g^{-1})s^{-1}g=p_3$, that means that $g^{-1}sg\in N_{g^{-1}Sg}(P)$. By other part, using the two left equalities, we have that 
$$gxg^{-1}(gp_1g^{-1})gx^{-1}g^{-1}=s(gp_2g^{-1})s^{-1}$$
that means that 
$$gxp_1x^{-1}g^{-1}=s(gp_2g^{-1})s^{-1} \Leftrightarrow xp_1x^{-1}=g^{-1}sgp_2g^{-1}s^{-1}g=p_3$$
and here is my problem. This means that $x\in N \Leftrightarrow x\in N_S(P)$?

A:

The set only states that $x\in N$ iff $c_g\circ c_x \circ c_{g^{-1}}\in \mbox{Aut}_S(gPg^{-1})$. This implies that $x\in N$ iff $c_{gxg^{-1}}=c_s$ for certain $c_s\in \mbox{Aut}_S(gPg^{-1})$, that means that $c_{s^{-1}gxg^{-1}}=1_{gPg^{-1}}$, that means that $s^{-1}gxg^{-1}$, but is not possible to say that $x\in N \Leftrightarrow x\in N_S(P)$.

