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Let $S$ be the dyadic solenoid.

Let $x\in S$, and let $X$ be the union of all arcs (homeomorphic copies of $[0,1]$) in $S$ containing $x$.

$X$ is called a composant of $S$.

It is well-known that $X$ is a dense first category one-to-one continuous image of the reals, and that $S$ is a homogeneous continuum.

Now let $y$ and $z$ be any two points in $S\setminus X$.

Is there a self-homeomorphism $h:S\to S$ such that $h[X]=X$ and $h(y)=z$?

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In this paper of J.Kwapisz I have found the following

Theorem 1. Any homeomorphism $h$ of the dyadic solenoid $S$ is isotopic to the "affine" homeomorphism of the form $g:x\mapsto \pm(2^n x+b)$ for some $n\in\mathbb Z$ and some $b\in S$.

If $h$ preserves the path-connected component $X$ of the neutral element, then so does the affine homeomorphism $g$, which implies that $b\in X$. It follows that for any $y\in S\setminus X$ the image $h(y)$ belongs to $g(y)+X\subset \pm 2^{\mathbb Z}y+X$, which is contained in a countable union of path-connected components and hence cannot be an arbitrary element $z\in S\setminus X$.

So, the answer to the original question is negative.

The same negative answer holds for homeomorphisms of any (not necessarily dyadic) solenoid.

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  • $\begingroup$ That is very surprising to me! $\endgroup$ – Forever Mozart Aug 15 '18 at 15:49
  • $\begingroup$ One typo: I think $h(x)$ should be $h(y)$ $\endgroup$ – Forever Mozart Aug 15 '18 at 15:57
  • $\begingroup$ @ForeverMozart The typo is corrected. Thanks for noticing. $\endgroup$ – Taras Banakh Aug 15 '18 at 16:10

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