$\DeclareMathOperator\bso{\beta^\*\!\omega}\DeclareMathOperator\Homeo{Homeo}$Let $\bso$ be the complement of the countable discrete space $\omega$ in its Stone-Čech compactification $\beta\omega$ (some authors denote it $\omega^\*$).



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> **Question.** Assume ZFC+CH. Does there exist a self-homeomorphism of $\bso$ with an isolated fixed point (i.e., a fixed point that is not a limit of other fixed points)?
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Bonus question: if yes: it it true that for every $x\in\bso$ there exist a self-homeomorphism of $\bso$ with $x$ an isolated fixed point?


Other bonus question: what about asking the self-homeomorphism to have order 2?




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Contextual remarks:


1. Say that a self-homeomorphism of $\bso$ is smooth if it is induced by a bijection between two subsets of $\omega$ with finite complement. If $f\in\Homeo(\bso)$ is smooth then the set of fixed points of $f$ is clopen, hence has no isolated point.
2. Shelah in the early 1980s showed the existence of models of ZFC in which every self-homeomorphism of $\bso$ is smooth (and in particular in which $\#(\Homeo(\bso))=\mathfrak{c}$). Hence in such models the above question has a negative answer.
3. Rudin proved in the 1950s that under ZFC+CH, $\#(\Homeo(\bso))=2^\mathfrak{c}$ and its action on $\bso$ is not transitive.
4. If there's a self-homeomorphism with an isolated fixed point $x$, it is easy to modify it to get another one with unique fixed point $x$.



Self-homeomorphism of Stone-Čech boundary with an isolated fixed point
The answer to your main question is *yes*. In fact, there is (under $\mathsf{CH}$) a self-homeomorphism of $\omega^\*$ with exactly one fixed point. Such a mapping is constructed in the proof of Theorem 5.7 in my paper:



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> "$P$-sets and minimal right ideals in $\mathbb N^\*$," *Fundamenta Mathematicae* **229** (2015), pp. 277-293. ([pdf](https://wrbrian.files.wordpress.com/2012/01/brian-4.pdf))
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*Sketch of the construction:* Begin with the shift map $\sigma$ on $\omega^\*$, which is the (unique) self-homeomorphism of $\omega^\*$ that is induced by the mapping $n \mapsto n+1$ on $\omega$. (That is, $\sigma$ maps every ultrafilter $\mathcal U$ to the ultrafilter generated by $\{ A+1 :\, A \in \mathcal U\}$.) This map has no fixed points. Using some ideas from earlier in the paper, there is a nowhere-dense, closed $P$-set $X \subseteq \omega^\*$ such that $X$ is closed under $\sigma$. (Recall that $X$ is a *$P$-set* if the intersection of any countably many neighborhoods of $X$ is again a neighborhood of $X$.) Now modify the topology of $\omega^\*$ by collapsing $X$ to a point. Assuming $\mathsf{CH}$, this quotient space is again homeomorphic to $\omega^\*$. ($\mathsf{CH}$ implies that collapsing a nowhere dense $P$-set to a point always results in a copy of $\omega^\*$; see Corollary 1.2.4 in van Mill's survey article about $\beta \omega$ for a proof.) The map $\sigma$ induces a self-homeomorphism of the quotient space, and the collapsed copy of $X$ is its unique fixed point. $\quad \square$


Notice that the construction is quite flexible. If you wanted exactly $17$ fixed points, this could be arranged. If you wanted no fixed points but a point of order $42$, then this could be arranged as well. (In the above argument, instead of collapsing $X$ to a point, collapse each of the $42$ sets $X \cap (k+42\mathbb N)^\*$ to a point.) I think the most general thing one could achieve along these lines is: for any function $f: \mathbb N \rightarrow \mathbb N \cup \{\aleph\_0\}$, there is (assuming $\mathsf{CH}$) a self-homeomorphism of $\omega^\*$ having $f(n)$ points of order $n$ for every $n$.


For the bonus questions, I think this should be possible using a different technique (but I'd have to check the details to be sure). Under $\mathsf{CH}$ I think that any point $u \in \omega^\*$ is a *bowtie point*, meaning that $\omega^\* \setminus u$ can be partitioned into two nonempty clopen sets. And I think these two sets are homeomorphic, so permuting them (while leaving $u$ fixed) gives you the kind of map you want. Note that this also gives you a different way of getting a positive answer to your main question as well.


