Let $G$ be a group, $D$ a $p$-subgroup of $G$. Suppose $x,y$ both lie in $C_G(D)$ the centraliser of $D$, that is, $xd=dx$ and $yd=dy$ for all $d\in D$.
I am solving a problem about the centre of the group algebra $kG$, which I have reduced to showing that: $x,y$ are conjugate in $G$, that is $\exists g\in G$ such that $gxg^{-1}=y$, if and only if $x,y$ are conjugate in $N_G(D)$, where $N_G(D)$ is the normaliser in $G$ of $D$.
It is clear that if $x,y$ are conjugate in $N_G(D)$ then they are conjugate in $G$. If $D$ is the trivial subgroup these two notions agree, but I'm not sure how to do more than this.
Plausibly useful things are that $N_G(D)$ acts on $C_G(D)$ by conjugation, but I can't get an argument about the size of the two conjugacy classes to work.
Any hints much appreciated, I'm sure I'm missing something obvious.
I am afraid that this is not true in general.
For a counterexample, let $G = A_6$, and let $P = \langle x,y \rangle \in {\rm Syl}_2(G)$, where we can take $x = (1,2)(3,4)$, $y = (3,5)(4,6)$.
Then $P$ is dihedral of order $8$, and we let $D := Z(P) = \langle (3,4)(5,6) \rangle$.
Then $C_G(D) = N_G(D) = P$, and $x$ and $y$ are conjugate in $G$ but not in $P$.