Let $X$ be a set and $u$ be a free ultrafilter on $X$. We can consider a topology on $X$ by declaring every element of $u \cup \{\emptyset \}$ to be open. 


El'kin's original motivation for looking at this topology is that $X$ is a space without isolated points which can't be split into disjoint dense subspaces. Another quirky feature: if $X$ is say $\mathbb{R}$ then translations are not continuous. This last fact makes me wonder: are there any periodic-point free continuous self maps at all on $X$ with this topology? In other words:



> 
> Is there a free ultrafilter $u$ on a set $X$ and a periodic point-free onto map $f: X \to X$ such that: $f^{-1}(A) \in u$, for every $A \in u$? Is there at least one such fixed-point free map?
> 
> 
> 



Periodic point-free maps and free ultrafilters.
Your question has been answered by Joseph with several references,
but since this nice result has several attractive proofs, let me
try to provide one.


**Theorem.** If $\mu$ is an ultrafilter on a set $X$ and
$f:X\to X$ has the property that $A\in\mu\leftrightarrow
f^{-1}A\in\mu$, then $f(x)=x$ for $\mu$-almost all $x$.


Proof. Consider the function $f$ as a directed graph, where each
point $x$ has an edge to $f(x)$.


Suppose first that $\mu$ happens to concentrate on the set $A$ of
points lying on a finite cycle. By the axiom of choice, let
$D\subset A$ be a maximal set of non-adjacent points in $A$. So
$D$ contains at least one of $a$, $f(a)$ and $f(f(a))$ for any
$a\in A$. It follows that $A\subset D\cup f^{-1}D\cup f^{-2}D$,
and so one of these sets must be in $\mu$. The main hypothesis
then implies that actually all of these sets are in $\mu$. But
notice that any point $y\in D\cap f^{-1}D$ must have $f(y)=y$,
since otherwise $y$ and $f(y)$ would be adjacent points in $D$. So
$\mu$ concentrates on fixed points of $f$, as desired.


To see that this is the only case, suppose next towards
contradiction that $\mu$ concentrates on the set $B$ of points
that are not yet on a cycle, but whose iterates eventually reach a
cycle. Let $B\_0$ be those points in $B$ that reach their cycle
first after an even number of iterations of $f$, and $B\_1$ the
points that do so first after an odd number of iterates. These
sets are disjoint, but $f^{-1}B\_0\subset B\_1$ and
$f^{-1}B\_1\subset B\_0$, and so actually neither can be in $\mu$,
contradicting $B=B\_0\sqcup B\_1\in\mu$.


Finally, assume toward contradiction that $\mu$ concentrates on
the set $C$ of points whose iterates are not eventually periodic.
This set is the union of those connected components of the graph
that do not contain a cycle. (Each such component is therefore a
tree.) By the axiom of choice, let $D$ select exactly one point
from each component of $C$. Let $C\_0$ be the points in $C$ whose
shortest distance to a point in $D$ has even length, and $C\_1$ the
points with odd distance to $D$. Thus, $C=C\_0\sqcup C\_1$ is a
partition of $C$, and $f^{-1}C\_1\subset C\_0$ and $f^{-1}C\_1\subset
C\_0$, since applying $f$ once will change the distance by exactly
one. So again, neither set can be in $\mu$, since either would
force the other also into $\mu$, contradicting that they are
disjoint.


So the only possible case is that $\mu$ concentrates on the fixed
points of $f$. QED


The theorem appears to be first proved by Katetov, Commentations
Math Univ Carolinae 8 (1967), 432-433, with a related result given
by Frolik 1968, and Andreas Blass's dissertation (1970).


You may also be interested in the [proof of the theorem written by
Bob Solovay](http://math.berkeley.edu/~solovay/Preprints/Rudin_Keisler.pdf).


Finally, I note that the proof uses the axiom of choice, and I am given to understand that the theorem is not provable in ZF. Perhaps someone else can post an answer explaining how a positive answer to your question is consistent with ZF, even though it is ruled out in ZFC. 


