Question
What functions on 2-adic space are in the conjugacy class of $f(x)=x+2^{\nu_2(x)}$?
A conjugacy class is the set of functions whose action conjugates to any other element of the class, in this case via a homeomorphism. Conjugacy, being reflexive, symmetric and transitive is an equivalence relation.
I'm looking for a classification which is at least exhaustive over functions of the form:
$f(x)=ax+b\cdot2^{\nu_2(x)}$
I can give some simple and more complex examples.
Example 1
$a=1,b=-1$ is an easy trivial example via the homeomorphism $x\mapsto -x$
Example 2
Moreover, examples varying $b$ are easy. 2-adic numbers can be represented with any radix $(0,d)$ where $\nu_2(d)=0$ so the case $b=d$ gives us all $\{(a,b):a=1,\nu_2(b)=0\}$
Example $3$
I know of examples for $a\neq1$. A corollary of Jyrki Lahtonen's answer here is that $3x+2^{\nu_2(x)}$ is conjugate to $x+2^{\nu_2(x)}$
If we take as a given that $\nu_2(b)=0$ then can we pick arbitrary $a$ such that $\nu_2(a)=0$ ?
Observation
Relevant to the above question is the fact that $f(x)$ commutes with $g(x)=2x$, i.e. $f(2x)=2f(x)$. This means for any $a,b$ which are conjugate, $2^m(a,b)$ is also conjugate. So one can pick either $a,b$ as satisfying $\nu_2=0$ and have a canonical representation of the class with $a,b$ drawn from $\Bbb Z_2^\times$. Essentially here I am asking a very similar question as what functions of the form $ax+b$ are conjugate to $x+1$.
Hypothesis
Is it the case that $f$ is conjugate for all $\{(a,b):a\in\Bbb Z_2^\times, b\in\Bbb Z_2^\times\}$?
Jyrki's answer to my previous linked question generalises to $f_{a,b}(x)=ax-b\cdot2^{\nu_2(x)}$ with $a,b$ any pair of $2$-adic units and therefore all these are topologically conjugate.
(Here $2^{\nu_2(x)}$ is the highest power of $2$ that divides $x$.)
Then if we let $f^n(x)$ indicate the $n^{th}$ composition of $f(x)$ we can define a 2-adic isometry $T$ as follows:
Let $\displaystyle T=\sum_{n=0}^\infty 2^{\nu_2(f^n(x))}$
And $T:\Bbb Q_2\to\Bbb Q_2$ is the isometry which topologically conjugates $f(x)$ to $g(x)=x-2^{\nu_2(x)}$ by satisfying the equation $T(f(T^{-1}(x)))=g(x)$
I will call $T$ the parity vector of $f$.
And there is a little more. Let $T_{a,b}$ be the parity vector associated with $f_{a,b}$ by the above method.
We can say $T_{1,1}(x)=g(x)$ is the identity of a group sitting as a subgroup of the homeomorphisms of $\Bbb Q_2$ because $T_{1,1}(x)=x$.
And for every $f_{1,b}(x)$ it is self-evident that $T$ simply exchanges the radix $(0,1)$ for $(0,b)$ which implies $f_{1,b}(x)=bx$ and therefore we have that the group inverse of $T_{1,b}(x)$ is $T_{1,b^{-1}}(x)$ and the group operation for $a=1$ is $T_{1,b_1\times b_2}(x)$.
Then as a minimum scope of the group we have $T_{1,b}():b\in\Bbb Z_2^\times$ and the group is the "scalings" of $\Bbb Q_2$ by $b\in\Bbb Z_2^\times$ although I'm aware scalings is probably the wrong word given that $b$ are units.
The perhaps less trivial case is the operation when $a\neq 1$ which may have implications for a range of Collatz-like problems. It looks likely $f_{a_1,b_1}(x)\circ f_{a_2,b_2}(x)=f_{a_1\times a_2,b_1\times b_2}(x)$.
I will move on to investigate whether $b$ has an inverse, and what it is, next.