I'm looking for an example of a group $G$ with two nonconjugate subgroups $H$, $K$ and two elements $a$, $b$ in $G$ such that $$H^a \lneq K \lneq H^b$$ where $H^g = g^{-1}Hg$ is the conjugate of $H$ by $g$. Such a group is necessarily infinite and nonabelian. Does anybody know an example?
This came up in the following context. The set $\mathscr{S}(G)$ of subgroups of $G$ is partially-ordered with respect to $\le$ (inclusion), and this relation descends to the set $\mathscr{C}(G) = \mathscr{S}(G)/G$ of conjugacy classes of subgroups by declaring $[H] \le [K]$ if and only if $H^g \le K$ for some $g$ in $G$. This is evidently reflexive (take $g = 1$) and it's transitive because $H \le K$ implies $H^g \le K^g$ for all $g$ in $G$. But anti-symmetry might not hold: if $[H] \le [K] \le [H]$ then $H^{g_1} \le K$ and $K^{g_2} \le H$, so $H^{g_1g_2} \le H$; if $G$ is finite, then comparing cardinalities we must have $H^{g_1g_2} = H$ so $H = K^{g_2}$; but if $G$ is infinite it's possible that $H^g < H$. For example, take $G = \mathfrak{S}_{\mathbf{Z}}$ and $H$ the subgroup of functions fixing the negative integers, and $\tau$ the translation $n \mapsto n - 1$. Then $H^\tau < H$ because every function in $H^\tau$ also fixes $0$. Still, it's not clear in this case that we can find a $K$ to squeeze in between $H^\tau$ and $H$, let alone a $K$ that is not ultimately conjugate to $H$.
Consider the group consisting of those matrices $M(n,r)=\begin{pmatrix}4^n & r\\ 0 & 1\end{pmatrix}$, for $(n,r)\in\mathbf{Z}\times\mathbf{Z}[1/2]$ (this is a semidirect product $\mathbf{Z}\times\mathbf{Z}[1/2]$, where the positive generator of the left-hand $\mathbf{Z}$ acts by multiplication by 4). Let $H_m$ be the cyclic subgroup generated by $M(0,m)$. Then $H_1\subset H_2\subset H_4$, $H_1$ and $H_4$ are conjugated (by $M(1,0)$), but $H_2$ is not conjugate to $H_1$.
It also has the Baumslag-Solitar presentation $\langle t,x\mid txt^{-1}=x^4\rangle$. Then $\langle x\rangle\subset \langle x^2\rangle\subset \langle x^4\rangle$, $\langle x\rangle$ and $\langle x^4\rangle$ are conjugate (by $t$) but are not conjugate to $\langle x^2\rangle$.