Let $hf=g$ and let $hfh^{-1}=f$ then what does this say about the relation between $f$, $g$ and $h$? I was studying composition of the functions:
$f(x)=2x+1$
$g(x)=\frac{2x-1}3$
$h(x)=\frac{x-2}3$
in a 2-adic setting but I guess the 2-adic part is largely irrelevant here and this applies more generally to groups.

Let $hf=g$ and let $hfh^{-1}=f$ then what does this say about the relation between $f$ and $g$?

My Attempt
My first observation is that the statement seems to say $f,h$ commute.
So I substituted $h$ and $h^{-1}$ into $hfh^{-1}=f$:
$h=gf^{-1}$
$h^{-1}=fg^{-1}$
$gf^{-1}ffg^{-1}=f$
giving me:
$gfg^{-1}=f$
So then by symmetry, I wondered if $ghg^{-1}=h$ as well and I was able to verify all elements commute with each other.  This doesn't appear to be a standard property of function composition. I have a tentative notion of a commutator group here or a conjugacy class but would appreciate a pointer that maps the elements of this structure on to some relevant literature. Coming back to the specific $f,g,h$ at the top (subject to domain / range), are there infinitely many other examples that commute with these functions?
I'm half expecting somebody to hit me with some linear algebra at this point showing the reflections of a triangle in its bisectors.
Just to add something on the 2-adic component of this, we have $f:\Bbb Z_2\to\Bbb Z_2^\times$ and $g:\Bbb Z_2\to\Bbb Z_2^\times$ but $h$ maps $\{x:\nu_2(x)=1\}$ to $\{x:\nu_2(x)>1\}$ and leaves the 2-adic value of everything else unchanged.
 A: I am not sure this is what you are asking for, still.
You are dealing with affine functions of the form
$$
f_{a, b}  : x \mapsto a x + b.
$$
Now it is immediately seen that $f_{a, b}$ and $F_{c, d}$ commute iff
$$
b (c - 1) = d (a - 1).
\tag{eq}
$$
If you start with $f = f_{2,1}$, you see that a function $f_{c, d}$ commutes with $f$ iff $c = d + 1$. It is then immediate to see that all these functions commute with each other.
One readily sees that in general the affine functions that commute with a fixed $f_{a, b}$ also commute among themselves, except of course when $a = 1, b = 0$, and $f_{1, 0}$ is the identity function.
Let us see that details. If $a = 1$, so that $b \ne 0$, we get from (eq) that $f_{c, d}$ commutes with $f_{a, b}$ iff $c = 1$, so the functions that commute with $f_{a, b}$ are the translations, which obviously commute pairwise.
Suppose then $a \ne 1$, and let $f_{c, d}$ and $f_{e, g}$ commute with $f_{a, b} \ne f_{1, 0}$. We get from (eq)
$$
b (c - 1) = d(a-1), \quad b(e-1) = g(a-1).
$$
Multiplying the first relation by $(e - 1)$, we get
$$
d (a - 1) (e - 1)
=
b (c - 1) (e - 1)
=
g (a - 1) (c - 1),
$$
so that $d ( e -1) = g (c - 1)$, as $a \ne 1$, and thus by (eq) $f_{c, d}$ and $f_{e, g}$ commute.

Addemdum
The above works for coefficients in a domain. Perhaps I should add that if we are working over a field $F$, we are dealing with the affine group $G = \operatorname{Aff}(1, F)$. If $g \in G$ is not a translation, then it has exactly one fixed point, and this has to be fixed by any element in the centraliser of $g$. Conversely, a $1$-point stabiliser is isomorphic to the corresponding general linear group, which in this case is (the abelian group) $F^{\star}$.
