TITLE: Normal subsets of a Sylow p subgroup are conjugate if and only if they are $N_G(P)$ conjugate.
QUESTION [9 upvotes]: The following is a question from Dummit & Foote. 

Prove that if $U$ and $W$ are normal subsets of a Sylow $p$-subgroup
  $P$ of a finite group $G$ then $U$ is $G$-conjugate to $W$ if and only
  if $U$ is $N_G(P)$-conjugate to $W$.

Ofcourse G-conjugate means, there exists a $g 
\in G$ such that $gUg^{-1}=W$, and for $N_G(P)$ conjugate the element is restricted to $N_G(P)$. 
The reverse implication is obvious. But I havent been able to prove the implication. D&F gives the hint that $N_P(U)=N_P(W)=P$, but I have not idea how this can be used to get information about $N_G(P)$. 
Any help will be appreciated. 
Thanks!

REPLY [9 votes]: Here's a sketch, there are a few details left out for you to fill in:

Assume $gUg^{-1} = W$ for some $g \in G$.
Show that $gPg^{-1} \subseteq N_G(W)$.
Use the Sylow theorems to get that $xgPg^{-1}x^{-1} = P$ for some $x \in N_G(W)$.
Observe that $xg \in N_G(P)$ and $xgUg^{-1}x^{-1} = W$.