In working on a paper, I found myself in the amusing but unusual situation of having a group $latex G$, a subgroup $latex H$ and an element $latex \xi\in G$ such that
$latex \xi H\xi =H.$
This certainly can happen: the two obvious cases are when $latex H=G$, or when $latex \xi$ is an involution that happens to be in the normalizer $latex N(H)$ of $latex H$.
In fact the general case is just a tweak of this last case: we have $latex \xi H\xi=H$ if and only if $latex \xi^2\in H$ and $latex \xi\in N(H)$, or in other words, if $latex \xi$ belongs to the normalizer, and is an involution modulo $latex H$.
This is of course easy to check. I then asked myself: what about the possibility that
$latex \xi^a H\xi^b = H$,
where $latex a$ and $latex b$ are arbitrary integers? Can one classify when this happens? The answer is another simple exercise (that I will probably use when I teach Algeba I next semester): this is the case if and only if $latex \xi^{a+b}\in H$ and $latex \xi^{(a,b)}\in N(H)$, where $latex (a,b)$ is the gcd of $latex a$ and $latex b$. In particular, for all pairs $latex (a,b)$ where $latex a$ and $latex b$ are coprime, the condition above implies that $latex \xi$ belongs to the normalizer of $latex H$.
Here is the brief argument: having fixed $latex \xi$, let
$latex M=\{(\alpha,\beta)\in\mathbf{Z}^2\,\mid\, \xi^{\alpha}H\xi^{\beta} =H\}.$
This set is easily seen to be a subgroup of $latex \mathbf{Z}^2$. Furthermore, note that $latex (\alpha,\beta)\in M$ implies that $latex \xi^{\alpha+\beta}\in H$, which in turns means that $latex (\alpha+\beta,0)\in M$ and $latex (0,\alpha+\beta)\in M$.
Hence if $latex (a,b)\in M$, we get
$latex (a,-a)=(a,b)-(0,a+b)\in M,\quad\quad (b,-b)=(a+b,0)-(a,b)\in M$,
so that
$latex M\cap (1,-1)\mathbf{Z}$
contains $latex (a,-a)\mathbf{Z}\cap (b,-b)\mathbf{Z}$, which is just
$latex (d,-d)\mathbf{Z},$
where $latex d=(a,b)$. But $latex (d,-d)\in M$ means exactly that $latex \xi^{d}\in N(H)$.
Thus we have got the first implication. Conversely, the conclusion means exactly that
$latex (a+b,0)\in M,\quad (d,-d)\in M.$
But then
$latex (a,b)=(a+b,0)-(b,-b)=(a+b,0)-b/d (d,-d)\in M$
shows that $latex \xi^a H\xi^b=H$.
To finish, how did I get to this situation? This can arise quite naturally as follows: one has a collection $latex X$ of representations $latex \rho$ of a fixed group $latex \Gamma$, and an action of a group latex $latex G$ on these representations $latex X$ (action up to isomorphism really).
For a given representation $latex \rho_0$, we can then define a group
$latex H=\{g\in G\,\mid\, g\cdot \rho_0\simeq \rho_0\}$
and also a subset
$latex T=\{g\in G\,\mid\, g\cdot \rho_0\simeq D(\rho_0)\},$
where $latex D(\cdot)$ denotes the contragredient representation.
It can be that $latex T$ is empty, but let us assume it is not. Then $latex T$ has two properties: (1) it is a coset of $latex H$ (because $latex H$ acts on $latex T$ simply transitively); (2) we have $latex g^2\in H$ for all $latex g\in T$ (because the contragredient of the contragredient is the representation itself).
This means that, for some $latex \xi\in G$, we have $latex T=\xi H$, and furthermore
$latex \xi H\xi=H$
since $latex \xi g\xi g\in H$ for all $latex g\in H$.
By the previous discussion, we therefore get a good handle on the structure of $latex T$: either it is empty, or it is of the form $latex \xi H$ for some $latex \xi\in G$ such that $latex \xi^2\in H$ and $latex \xi$ normalizes $latex H$ in $latex G$. In particular, if $latex H$ is trivial (which happens often), either $latex T$ is empty, or it consists of a single element $latex \xi$ which is an involution of $latex G$.
