TITLE: If $H$ and $K$ are conjugate in $G$, are they conjugate in
QUESTION [7 upvotes]: Suppose $G$ is a group, with $H$ a subgroup.  Suppose that $K$ and $L$ are subgroups in $H$ that are conjugate in $G$, so we have an element $g\in G$ with $gKg^{-1}=L$.  Does it follow that $K$ and $L$ are conjugate in $N_G(H)$?
Since $G=N_G(H)$ if $H$ is normal, the result holds for normal $H$.  If the result is not true in general, are there any less strict conditions (besides normality of $H$) to place on $(G,H,K,L)$ so that it is true?

REPLY [4 votes]: In technical terms you are asking if “$H$ controls its own fusion”. If $H$ is an abelian Sylow $p$-subgroup, or more generally if $K,L$ are normal subgroups of the Sylow $p$-subgroup $H$, then the answer is yes (Burnside's fusion theorem).
In general, no. Consider $K=\langle(1,2)(3,4)\rangle$, $L=\langle(1,3)(2,4)\rangle$, $H=\langle(1,2), (3,4), (1,3)(2,4) \rangle$, $G=\langle (1,2), (1,2,3,4) \rangle = S_4$.
$K$ and $L$ are conjugate in $N_G(\langle K,L\rangle) = N_G(K_4) = S_4$, but not in $N_G(H) = H = D_8$.
Subgroups of the form $N_G(P)$ for $P \leq H$ ($H$ a Sylow $p$-subgroup) are called $p$-local. $p$-local subgroups control fusion (Alperin's theorem), but you sometimes need more than one of them.