Are $f(x)=2x+1$ and $g(x)=\frac{2x-1}3$ topologically conjugate on $\Bbb Z_2$? Let $f,g$ be functions from $\Bbb Z_2\to\Bbb Z_2^\times$
$f(x)=2x+1$ and $g(x)=\frac{2x-1}3$
Then I have the notion that $f,g$ are topologically conjugate to each other in the 2-adic metric.  I'd like to verify that but I find it offputting that the domains of the functions don't coincide with the range.
Anyway, proceding regardless, I can ask myself "is there a homeomorphism that conjugates one to the other?"
So I guess the homeomorphism would be $x\mapsto\frac{x-2}3$ or $x\mapsto3x+2$
So to my tiny brain, that homeomorphism looks good because it's one to one and onto (because the maps to and from zero fall outside of $\Bbb Z_2^\times$, preserves the 2-adic value $\lvert x\rvert_2=1$ and $d(f(x),f(y))=d(x,y)$.  Is that correct?
I'm concerned I don't have any conjugation going on there, just a transformation from one to the other.  But how could I, as the codomain is not within the range?
EDIT:
I've moved my thinking along.  If I let $h=\frac{x-2}3$ and conjugate $f$ I seem to get $hfh^{-1}=f$ rather than $g$ so I'm definitely doing something wrong here. Note, I have no doubt they are topological conjugates, but I want to derive the homeomorphism.
The fact $hfh^{-1}=f$ suggests $f,h$ commute - so i checked whether $ax+b$ and $cx+d$ commute in general, and I got $hfh^{-1}=ax+(1-a)d+bc$, confirming they DON'T in general commute, but I substituted in this special case and got $ax+(1-a)d+bc=2x+1$.  I found this which looks relevant but I seem to be going down a rabbit hole not relevant to the original question.
 A: It turns out that $f$ and $g$ actually are topologically conjugate.
Let's simplify the question first a bit by conjugating the two functions with the translation $t(x)=x+1$. We get
$$
\tilde{f}(x):=t(f(t^{-1}(x)))=(2(x-1)+1)+1=2x,
$$
and
$$
\tilde{g}(x):=t(g(t^{-1}(x)))=([2(x-1)-1]/3)+1=2x/3.
$$
Topological equivalence is an equivalence relation, so $f$ and $g$ are conjugate if and only if $\tilde{f}$ and $\tilde{g}$ are.
Let us then define the function $h:\Bbb{Z}_2\to\Bbb{Z}_2$ by the recipe
$$
h(x)=\begin{cases}0,&\text{if $x=0$, and}\\
3^{\nu_2(x)}x,&\text{otherwise.}\end{cases}
$$
Here, for all $x\neq0$, $\nu_2(x)=\ell$, when $x\in 2^\ell\Bbb{Z}_2\setminus2^{\ell+1}\Bbb{Z}_2$. Then

*

*As $3$ is a $2$-adic unit, and $\Bbb{Z}_2$ is the disjoint union of the subsets $U_\infty:=\{0\}$ and $U_\ell:=2^{\ell}\Bbb{Z}_2\setminus 2^{\ell+1} \Bbb{Z}_2$, $\ell=0,1,2,\ldots$, it follows that $h$ maps each subset $U_\ell, \ell=0,1,\ldots,\infty$,
bijectively onto itself. Therefore $h$ is itself also a bijection.

*If $V=x+2^\ell\Bbb{Z}_2$ is any basic open subset of $\Bbb{Z}_2$, then
$$h^{-1}(V)=h^{-1}(x)+2^\ell\Bbb{Z}_2$$ is another basic open subset. Therefore $h$ is continuous.

*The argument of the previous bullet holds equally for the obvious inverse of $h$ (replace $3$ by $1/3$ everywhere), so we can conclude that $h$ is a homeomorphism.

*Finally, we also have, for all $x\neq0$ in $\Bbb{Z}_2$
$$h(\tilde{g}(x))=3^{\nu_2(2x/3)}(2x/3)=3^{1+\nu_2(x)}(2x/3)=3^{\nu_2(x)}\cdot 2x=\tilde{f}(h(x)).$$ So $h$ gives the topological conjugacy between $\tilde{f}$ and $\tilde{g}$ proving the main claim also.

