If $H$ and $K$ are conjugate in $G$, are they conjugate in 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?
 A: 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.
A: No, this does not follow. I will give a counter-example. I am sure counter-examples exist in finite groups, but my counter-example uses infinite groups ('cause it is dead easy in this world).
Take $G=F(a, b)$, the free group on the letters $a$ and $b$, and take $L=\langle a\rangle$ and $K=\langle b^{-1}ab\rangle$. Take $H=\langle a, b^{-1}ab\rangle$. We shall prove that $b\not\in N_H(G)$, which is sufficient (why?). To see that $b\not\in N_H(G)$, suppose otherwise. Then $b^{-2}ab^2\in H$. However, any freely reduced word over $a$ and $b^{-1}ab$ cannot contain a proper power of $b$. So we are done.
I think the condition you need is that if $\phi$ is an inner automorphism of $G$ then when $\phi$ restricts to $N_G(H)$ you have that $\phi$ is inner (which is subtly different from what Tobias wrote in the comments - note that not all automorphisms of $N_H(G)$ need be induced by an automorphism of $G$).
