The relation $x \sim g x g$ in groups While thinking about item (2) in Standard or good names for relations between maps, I thought I'd look at the relation $x \sim g x g$ in groups.
Going through all small groups of order at most 64, it seems to me, that for any finite group the connected components all have the same size, and that the number of connected components is a power of 2.  Is this obvious, or false?
NB: What I am really after is a name for the relation.
(edited to reflect YCor's observation)
(edited to reflect LSpice's observation)
 A: EDIT (11/01/2023) : I've expanded this answer to show that, if $G$ has a normal subgroup $N$ of odd order such that $G/N$ is an elementary abelian $2$-group, then the relation is universal on each coset of $N$.
(In particular, this covers the cases when $G$ has  odd or twice odd order, which were mentioned in the comments.)
Lemma: If $a$ and $b$ have odd order, then $aba=b$ implies $a=1$.
Proof: $aba=b$ implies $b$ inverts $a$ by conjugation. This means $b^2$ commutes with $a$ but since $b$ has odd order, $b$ commutes with $a$ so the original equation simplifies to $a^2=1$ which implies $a=1$ since $a$ has odd order.
Now, let $x\in G$ and let $\sigma_x:Nx\to Nx$ be given by $g\mapsto gxg$.
We show that $\sigma_x$ is injective. Let $g,h\in Nx$ such that $\sigma_x(g)=\sigma_x(h)$, that is $hxh=gxg=hh^{-1}gxhh^{-1}g$ which implies $xh=h^{-1}gxhh^{-1}g$. Writing $a=h^{-1}g$ and $b=xh$, this becomes $b=aba$. Since $G/N$ is an elementary abelian $2$-group and $x,g,h\in Nx$, it follows that $a,b\in N$ and the Lemma implies that $a=1$, that is $g=h$, as required.
By finiteness, $\sigma_x$ is also surjective and it follows that the relation is universal on $Nx$.
A: It's indeed quite immediate.
Indeed, let $\simeq$ be the equivalence relation generated by this relation. Then $x\simeq y$ iff the images of $x$ and $y$ in $G/G^2$ are equal. Here $G^2$ is the subgroup of $G$ generated by squares, so $G/G^2$ is the largest 2-elementary abelian quotient of $G$. ($G$ is not assumed finite.)
To prove the claim, one direction is clear ($x\simeq y$ implies $xG^2=yG^2$). For the converse, first observe that
$$x\simeq gxg\simeq hgxgh\simeq (hg)^{-1}hgxgh(hg)^{-1}=xghg^{-1}h^{-1},$$
and by induction it follows that $x\simeq xz$ for every $z\in [G,G]$. Then
$$xg^2=gxg\; g^{-1}(x^{-1}g^{-1}xg)g\simeq gxg\simeq x.$$
So by induction $xz\simeq x$ for every $z\in G^2$.
(For a name: it's the equivalence relation induced by equality modulo a normal subgroup, or equivalently, fibers of a group homomorphism, so this is the most standard kind of equivalence relations in group theory.)
