TITLE: The Stone-Čech compactification of the fixed point set
QUESTION [6 upvotes]: Let $G$ be a discrete group and $X$ be a Tychonoff $G$-space. Then there
exists a $G$-action on Stone-Čech compactification $\beta X$. If the
fixed point set $X^{G}\neq \emptyset $, then the Stone-Čech
compactification of the fixed point set is the fixed point set of the Stone-Čech compactification, i.e., $\beta \left( X^{G}\right) =\left( \beta
X\right) ^{G}$?

REPLY [5 votes]: Not without further assumptions.
First create an ordered space $X$ by identifying $\langle0,\omega_1\rangle$ and $\langle1,\omega_1\rangle$ in the product $2\times(\omega_1+1)$ to a point, $\Omega$ say. Then make the product $X\times X$ and take out $\langle\Omega,\Omega\rangle$. You can rotate the resulting space $Y$ around that hole over $\pi$ by mapping $\bigl<\langle i,\alpha\rangle,\langle j,\beta\rangle\bigr>$ to $\bigl<\langle 1-i,\alpha\rangle,\langle 1-j,\beta\rangle\bigr>$ and similarly for points with $\Omega$ as a coordinate. This map $f:Y\to Y$ has no fixed points but $\beta Y=X$ and $\beta f$ does have a fixed point: $\langle\Omega,\Omega\rangle$.
Now take the sum $Y\cup\{0\}$ and extend $f$ by $f(0)=0$. Then $f$ has one fixed point and $\beta f$ has two. As $f$ is its own inverse this gives us an action of $\mathbb{Z}/2\mathbb{Z}$ that is a counterexample.