Is $xSx^{-1} \subseteq S$ equivalent to $xSx^{-1} =S$ in a group? Let $G$ be a group and $\varnothing ≠S \subseteq G$ and given an element $x \in G$ I want to prove if the following implication holds (or disprove, if it doesn't):
$xSx^{-1} \subseteq S \iff xSx^{-1} = S$
My attempt: ($\Leftarrow$) is trivial, so I'll try to prove ($\Rightarrow$).
It is sufficient to show that $S \subseteq xSx^{-1}$, so I'll try prove by contradiction. Assume $S \nsubseteq xSx^{-1} \Rightarrow \exists s \in S: s \notin xSx^{-1}$ and hence $xsx^{-1} \in xSx^{-1} \subseteq S$, so $xsx^{-1} =s'$ for some $s' ∈ S \Rightarrow s =x^{-1}s'x \Rightarrow s \in x^{-1}Sx$
But now I'm not sure how to produce a contradiction. Any help?
 A: Counterexamples can be constructed with the concept of an "ascending HNN extension". The general HNN extension is defined by taking a group $G$, two subgroups $A,B < G$, and an isomorphism $f : A \to B$, choosing a presentation for $G$, and defining $G *_f$ to be the group defined by starting with the presentation for $G$, adding one generator $t$ called the "stable letter", and adding an extra defining relation $t a t^{-1} = f(a)$ for each $a \in A$. One proves that the natural homomorphism $G \mapsto G*_f$ is injective on $G$ and so also on $A$ and $B$.
In the group $G*_f$ it follows that $A$ is conjugate to $B$, because $tAt^{-1}=B$.
So to get a counterexample that disproves your implication, we simply need a group $G$ and two subgroups $A,B$ such that $A$ is isomorphic to one of its own proper subgroups $B < A$. In this situation the HNN extension that one gets by choosing an isomorphism $f : A \to B$ is called an ascending HNN extension.
Perhaps the very easiest example is $G = A = \mathbb Z$ and $B = 2 \mathbb Z$, with isomorphism $f : \mathbb Z \to 2 \mathbb Z$ defined by $f(1)=2$. One obtains the famous Baumslag-Solitar group
$$\langle a,t \mid tat^{-1}=a^2\rangle
$$
in which the infinite cyclic subgroup $\langle a \rangle$ is conjugate to its own index 2 subgroup $\langle a^2 \rangle$.
A: *

*It is possible to have an automorphism $\phi$ of $G$ and a subgroup $S$ such that $\phi(S) \subset S$, but $\phi(S) \ne S$.


*Given a group $G$, there exists a larger group $\tilde G$ ( the holomorph of $G$) that contains $G$ and $\phi$, and such that the action of $\phi$ on $G$ is given by conjugation ( see link for details). In fact, we can use even a subgroup of the holomorph.


*An example in 1. is $G=\mathbb{Q}$, $S= \mathbb{Z}$, and $\phi(x) = 2x$. Now the holomorph of $\mathbb{Q}$ is the group of the affine transformation of the rational line
$$f_{b,a}\colon x \mapsto a x + b$$
that can also be written as
$$\mathbb{Q} \rtimes \mathbb{Q}^{\times}$$
Inside this group we have $S = \mathbb{Z} \rtimes{1}$, the element $x=(0, 2)$, and $x S x^{-1} = 2 \mathbb{Z}$.
