Are two subgroups that contain a common element conjugate iff they are conjugate under the normalizer? I have two isomorphic subgroups $U,V\leq G$ and an element $a \in U \cap V$. Is it true that $U$ and $V$ are conjugate under $G$ if and only if they are conjugate under $N_G(\langle a \rangle)$? I vaguely remember hearing such a statement but at the moment I can neither think of a proof nor find a counterexample. Thanks in advance for any help.
 A: (To answer a question in the comments with an easy example)
This is not even true when U is abelian.  Conjugation of p-subgroups is controlled by local subgroups, but you cannot be quite so picky about which local subgroups you use (in particular, $N_G(\langle a\rangle)$ may be quite small).
Let $G=\operatorname{AGL}(1,8) \cong 7 \ltimes (2\times2\times2)$ be the group of invertible affine transformations $x\mapsto ax+b$ for $a,x,b$ in GF(8) and $a≠0$.  There are many Klein four-subgroups of G, but their normalizers are all the (normal) Sylow 2-subgroup P of order 8, $\{ x \mapsto x+b : b \in \operatorname{GF}(8)\}$.  For instance the subgroup U generated by $x\mapsto x+1$ and $x\mapsto x+\zeta$ for $\zeta$ a primitive 7th root of unity is conjugate in the whole group G to the subgroup V generated by $x \mapsto x+1$ and $x\mapsto x+\zeta^2$.  However, the normalizer of $x\mapsto x+1$ is the Sylow 2-subgroup P, and the two subgroups are obviously not conjugate inside an abelian group P.
In particular, the local subgroups are the Sylow 2-subgroup P as well as the entire group G.  The normalizer of every proper non-identity subgroup of P is P itself (since $G/P$ acts faithfully and irreducibly on P and is cyclic of prime order), so most local subgroups are P.  However, the normalizer of P itself is all of G and so it is this local subgroup that really controls the fusion.
Similar examples work for any prime p and elementary abelian Sylow p-subgroup P with proper subgroups U and $U^g$ that have P as the normalizer of their intersection (or any non-identity subgroup of their intersection).
For abelian Sylow p-subgroups, the normalizer of the Sylow always controls the fusion (Burnside's theorem), but the normalizer of proper p-subgroups may be much too small to control anything.
A: No, this is false as stated. For example, let $G = {\rm GL}(3,2).$ Let $a$ be an involution of $G.$ Let $U = C_{G}(a),$ which is dihedral of order $8.$ Let $V$ be another Sylow $2$-sbgroup of $G$ in which $a$ is not central. (This is possible, because all involutions in $G$ are conjugate, but not all involutions of $U$ are in $Z(U)).$ Now $U$ and $V$ are certainly conjugate within $G,$ as both are Sylow $2$-subgroups of $G.$ They are certainly not conjugate via an element of $N_{G}(\langle a \rangle) = C_{G}(a) = U.$
