I'm getting a contradiction I can't resolve - maybe you can help me?
Question
Does a solenoidal 2-adic function conjugate, by binary string reversal, to a continuous function on $\Bbb R$?
Problem
I'm getting that it both does and it doesn't.
A 2-adic function $\phi$ is solenoidal if it induces a function $\pmod{2^n}$, i.e. if:
$x\equiv y\pmod 2^n\implies f(x)\equiv f(x)\pmod{2^n}$
A class of functions $\phi$ act by conjugation on affine functions of the form $f_{a,b}(x):x\mapsto ax-b\cdot2^{\nu_2(x)}$. In fact for every pair $(a_1,b_1),(a_2,b_2)$ there is a solenoidal isometry which topologically conjugates $\phi_{a_2,b_2}$ to $\phi_{a_1,b_1}$. (In fact there are two, but by setting $\phi(0)=0$ we select one.
Let $\phi:\Bbb Z_2^\times\to\Bbb Z_2^\times$ be restricted to the units, and let this be conjugated to the dyadic rationals in the right hand half unit interval by reversal of its binary string, so e.g. $\overline{0}1_2$ would become $\frac12$ and $\overline{01}_2$ would become $\frac23$.
I am satisfied that if $\phi$ is further conjugated to the Cantor set, by replacing the ones with twos, this is a continuous function on a subset of $\Bbb R$. This is weakly confirmed by the following fact (i.e. not contradicted, when a contradiction is likely): A 2-adic isometry has periodic points only of order a power of two, and that yields a set compatible with Sharkovskii's theorem.
But what about the function on $[\frac12,1)$ written with ones.
Here's why it's continuous: If $f$ is solenoidal, then $f'(x):\Bbb R\to\Bbb R$ will agree on the first $n$ binary digits when $x$ agrees on the first $n$ binary digits, therefore it's continuous on $\Bbb Z[\frac12]\cap[\frac1{2}1)$ which is a dense subset so it's continuous.
Here's why it's discontinuous. The conjugation operation will overlay every negative natural number (as a subset of $\Bbb Z_2$) with a positive one, due to the equivalence of $\overline{1}0_2\sim\overline{0}1_2$ in $\Bbb R$. So even if we delete the right-limit dyadic rationals in $\Bbb R$ (the image of the negative natural numbers in $\Bbb Z_2$) or delete the image of the natural numbers (the left-limits of the dyadic rationals in $\Bbb R$), we'll have numbers arbitrarily close together whose preimages were apart.
I can't work out how to break either of the conflicting arguments.
Edit
Reflecting a little more, this definitely comes down to left-limits in $\Bbb R$ which converge to the same points as right-limits. For example the binary sequences in $\Bbb R$ $101,1011,10111,\ldots\to\frac34$ and
$111,1101,11001,110001,\ldots\to\frac34$
converge to different values when conjugated to $\Bbb Z_2^\times$ where they go to $-3$ and $3$ respectively (although they don't in general go to $x,-x$ so this function can't be confinuous.
$\frac34-2^{-n}:n\to\infty$
$\frac34+2^{-n}:n\to\infty$
Edit 2
Ok I worked a little more out. Let $s(x)$ be the conjugating function then the above means $s^{-1}$ isn't an open function therefore $s$ isn't continuous. I need to think what that implies for $s\phi s^{-1}$ but I'm tentatively thinking it means it's not continuous, because $s\phi$ is continuous and $s^{-1}$ isn't therefore $s\phi s^{-1}$ can't be.
Ok I worked this out. As I alluded to in the question, we would need to delete either the negative or the positive natural numbers as a subset of $\Bbb Z_2$ to make an injection into $[\frac12,1)$. But even then, suppose we delete the negatives, a sequence of positive natural numbers can still be made which converges to any given negative natural number in the 2-adic topology, for example:
$1_2,101_2,1101_2,11101_2\ldots$ or $\displaystyle\lim_{n\to\infty}2^n-3$
And the image of this in $[\frac12,1)$ will be a right-open set which converges to the limit of an equivalent left-open set.
Since $\phi$ is a bijection and both spaces are Hausdorff, $f'$ cannot be continuous. These two sequences which converge in $[\frac12,1)$ but whose reverse sequences have distinct limits in $\Bbb Z_2$ are:
$\frac34-2^{-n}:n\to\infty$
$\frac34+2^{-n}:n\to\infty$