Conjugacy in Infinite Groups 1) A theorem of Higman-Neumann-Neumann implies that any group $G$ can be embedded in a group $\tilde{G}$ such that any two isomorphic subgroups of $G$ are conjugate in $\tilde{G}$.
2) It is well known that the infinite cyclic group has only two automorphisms: identity and inversion.

Let us apply (1) for the infinite cyclic group $\langle x\rangle \cong \mathbb{Z}$. Then $\langle x\rangle$ and $\langle x^2\rangle$ are conjugate in some group $\tilde{G}$. Hence there is some $g\in \tilde{G}$ such that $g^{-1}xg=x^{2i}$ for some $i\neq 0$. This means, $g$ normalizes the subgroup $\langle x\rangle$, and induces an automorphism of $\langle x\rangle$, which takes $x$ to $x^{2i}$, which is different from $x,x^{-1}$. 
I arrived at some wrong conclusion by (2).
Please clarify the argument, and theorems, if anything wrong is there.
Thanks in advance!
 A: Indeed, it is possible for a subgroup to properly contain one of its conjugates. That is, it is perfectly possible for $gHg^{-1}\subsetneq H$. The normalizer $N_G(H)$ is the set of all $g$ such that $gH=Hg$, which is a stirctly stronger property than $gHg^{-1}\subseteq H$ in general (with say finite groups they are equivalent though, because conjugation is injective and we can check orders etc). Thus $gHg^{-1}\subset H$ does not tell us that $g$ normalizes $H$, and conjugation by $g$ does not restrict to an automorphism of $H$.
For the record, I take "$g$ normalizes $H$" to be equivalent to $g\in N_G(H)$, where $N_G(H)$ is the normalizer of $H$ in $G$ defined as the set of all $g\in G$ such that $gH=Hg$ (strict equality); this definition is consonant with that listed in GP, WP, MW, PM and presumably all references cited therein. I haven't seen $gHg^{-1}\subseteq H$ as the definition for "$g$ normalizes $H$" outside of any context that wasn't explicitly specifically for finite groups, if I recall correctly.
