Let $S$ be the dyadic [solenoid](https://en.wikipedia.org/wiki/Solenoid_(mathematics)).


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 $h(y)=z$?



homeomorphisms induced by composant rotations in the solenoid
In [this paper of J.Kwapisz](http://www.math.montana.edu/jarek/documents/papers/solenoid.pdf) 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.


