Uniqueness of conjugates of a subgroup. This question is partly influenced by the question:
Are two subgroups that contain a common element conjugate iff they are conjugate under the normalizer? 
If we have an arbitrary finite group $G$ and radical $p$-subgroups $A,B\leq G$ (so $A=O_{p}(N_{G}(A),p)$, $B=O_{p}(N_{G}(B),p)$ and the normalizers of $A$ and $B$ in $G$ are thus parabolic subgroups) such that $A\leq B\leq N_{G}(B)\leq N_{G}(A)\leq G$ (where $p\in\pi(G)$), is it the case that there is a unique conjugate of $A$ contained in $B$?
Clearly $A$ is normal in $B$, but I cannot seem to prove that there cannot be an elemnet $g\in G\setminus N_{G}(A)$ such that $A^{g}\ne A$ and $A^{g}\leq B$. Equally I have not found an example where uniqueness does not hold. Any hints would be greatly appreciated.
Edit:
Initially the question did not state that $A$ and $B$ were radical subgroups, as a greater understanding of the more general situation as deemed to be desirable.
 A: It is true for groups of Lie type in defining characteristic
A unipotent subgroup of a connected, reductive linear algebraic group is called radical if it is the unipotent radical of its normalizer. One version of the Borel–Tits theorem is that the radical subgroups are the unipotent radicals of the parabolic subgroups. Hence the radical subgroups of a fixed Borel subgroup are indexed by subsets of the positive roots, and so we have the rather convenient $A \leq B \implies N_G(B) \leq N_G(A)$ whenever $A,B$ are radical subgroups of a connected, reductive linear algebraic group.  Alperin's fusion theorem (for unipotent subgroups of a connected, reductive linear algebraic group) shows that whenever $\langle A,A^g \rangle$ is unipotent, there are radical subgroups $B_i$ and elements $g_i \in N_G(B_i)$ such that $A^{g_1 \cdots g_{i-1}} \leq B_i$ and $g = g_1 \cdots g_n$. In particular, if $A$ itself is radical and $\langle A,A^g \rangle$ is unipotent, then $A=A^g$ since all the $g_i \in N_G(B_i) \leq N_G(A)$. In other words, every radical subgroup of a connected, reductive linear algebraic group is weakly closed in every maximal unipotent subgroup containing it.
It is false in general
However, when $G$ has a characteristic different from $p$ (or has no characteristic), then things loosen up considerably. We no longer have $A \leq B \implies N_G(B) \leq N_G(A)$ for radical subgroups, and this allows us to focus attention on the “wrong” $B$, while another $B$ takes care of the conjugation. $\newcommand{\Sym}{\operatorname{Sym}}\newcommand{\PSL}{\operatorname{PSL}}$
A few infinite families of counterexamples for $p=2$ include $\Sym(n)$ for at least half of $n\geq 7$, and $\operatorname{PSL}(3,q)$ for $|q| \geq 5$ odd. I assume counterexamples are quite common in general, but there appear to be a lot of “small exceptions” to my assumption, so beware.
For $G=\Sym(7)$, let $A=\langle (1,2) \rangle$, which has order 2 and normalizer $A \times \Sym(\{3,4,5,6,7\})$ and let $B=\langle (1,2), (3,4), (5,6), (3,5)(4,6) \rangle$ be a self-normalizing Sylow 2-subgroup. Clearly $A$ has many $G$-conjugates in $B$, including $\langle (3,4) \rangle$ and $\langle (5,6) \rangle$. This does not contradict Alperin's fusion theorem, since the conjugation that takes $A$ to $\langle (5,6) \rangle$ does not have to rely  only on the subgroup $B$. A similar construction works for most $n$: if $2,3 \equiv n \mod 4$ then more or less the exact construction works. I am a little unclear on how easily $\Sym(9)$ works while $\Sym(8)$ does not work at all.
For $G=\PSL(3,q)$, ($|q| \geq 5$ an odd prime power, $\PSL(3,-|q|) = \operatorname{PSU}(3,|q|)$ being the unitary group), there are two cases for the proof, but the choice of $A$ and $B$ is the “same.”  Take $B$ to be a Sylow 2-subgroup and $A=Z(B)$ to be its center. For $q=1 \mod 4$, $B$ is a “wreathed” Sylow 2-subgroup, and $A=Z(B)$ is the “diagonal” subgroup of the base of $B$. $A$ is $G$-conjugate to all the direct factors of the base, but obviously is normal in $N_G(B)$.  Again this does not contradict Alperin's fusion theorem, since $A$ is contained in the base of $B$, a radical subgroup whose normalizer does not normalize $A$.  For $3 \equiv q \mod 4$, $B$ is quasi-dihedral and $A=Z(B)$ has order 2. $A$ is contained in a radical Klein 4-subgroup with full fusion, but any larger radical subgroup has smaller normalizer. I'm not entirely certain I understand why $q=3$ is an exception.
