Control of fusion in a sylow p subgroup

Let $G$ be a finite group, $p$ a prime, and $P\in {\mathrm{Syl}}_p(G)$ a Sylow $p$-subgroup (Sylow $p$-subgroups of a finite group). Following the convention $x^g=gx{g}^{-1}$ of The conjugacy class ${\mathrm{Cl}}_G(x)$ and centralizer $C_G(x)$ of an element, we say that

$P\text{ controls fusion in }P\text{ with respect to }G$

when, for all $x,y\in P$, the existence of $g\in G$ with $y=x^g$ implies the existence of $u\in P$ with $y=x^u$. Equivalently: any two elements of $P$ that are $G$-conjugate are already conjugate by an element of $P$; equivalently, the conjugacy class of $x$ in $G$ meets $P$ in exactly the conjugacy class of $x$ under $P$ (Subgroup, The normalizer $N_G(H)=\{g\in G:gH{g}^{-1}=H\}$ of a subgroup).

What is and is not claimed. This is a statement about element fusion only: it says nothing about when two subgroups of $P$ are conjugate in $G$, and nothing about elements of $G$ outside $P$. It is a property of the pair $(G,P)$, and it is invariant under conjugating $P$: if $g\in G$ and the Sylow $P$ controls fusion in $P$ with respect to $G$, then $P^g=gP{g}^{-1}$ controls fusion in $P^g$ with respect to $G$, because $y=x^u$ implies $y^g=(x^g{)}^{gu{g}^{-1}}$ and $gu{g}^{-1}\in P^g$ whenever $u\in P$. Control by $P$ implies control by $N_G(P)$, since $P\le N_G(P)$: every conjugating element supplied inside $P$ also belongs to $N_G(P)$. When $N_G(P)=P$ the two conditions are identical, but this equality alone does not assert that either condition holds.