Stabiliser of a Subset of Center I have no clue for the following problem: 

Let $G$ be a finite group, $p$ a prime number, $S$ a Sylow $p$ subgroup of $G$. Let $N$ be the normalizer of $S$ inside $G$. Let $X, Y$ two subsets of $Z(S)$ (center of $S$) such that $\exists g \in G, gXg^{-1}= Y$. Then we need to show that $\exists n \in N$ such that $gxg^{-1} = nxn^{-1}, \forall x \in X$. 

So I guess first I can assume $X, Y$ to be subgroups by taking the smallest subgroup containing them. Then I have no clue. 

I upvoted Arturo's answer, but I'm going to write out a more complete answer because I was also struggling with this question.
We'll follow Arturo's advice, and make the changes Derek Holt suggests in the comments.
$S$ centralizes $X$, so $gSg^{-1}$ centralizes $Y=gXg^{-1}$. However $S$ also centralizes $Y$. Thus $S$ and $gSg^{-1}$ are both Sylow $p$-subgroups of $C(Y)$. Thus there is $c\in C(Y)$ with $S=cgSg^{-1}c^{-1}$, since Sylow $p$-subgroups are conjugate. Since $S=(cg)S(cg)^{-1}$, $cg\in N(S)$. Then let $n=cg$, $x\in X$, and $y=gxg^{-1}$. Then we have
$$nxn^{-1}=cgx(cg)^{-1} = cgxg^{-1}c^{-1} = cyc^{-1}=cc^{-1}y=y=gxg^{-1},$$
as desired.
