Periodic point-free maps and free ultrafilters. 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?

 A: 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.
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. 
A: Although the question has already been thoroughly answered, it might be worthwhile to point out that the result can be separated into the combinatorial "meat", which doesn't involve ultrafilters, and a small corollary where the meat is fed to the ultrafilter problem.  The meat is the following theorem, which is, if I remember correctly, explicit in the early references; it is essentially proved (though not stated) in Joel Hamkins's answer here.  Given any set $X$ and any function $f:X\to X$, there is a partition of $X$ into four disjoint (possibly empty) sets $X=A_0\sqcup A_1\sqcup A_2\sqcup A_3$ such that $A_0$ is the set of fixed points of $f$ and each of the other three $A_i$'s is disjoint from its image under $f$.  Once one has this result, one immediately sees that any ultrafilter on $X$ must contain one of the four $A_i$'s and if that $A_i$ isn't $A_0$ then the ultrafilter can't be $f$-invariant as in the question.
A: Such a map does not exist since a continuous map on $X$ must be constant almost everywhere. In other words, if $f:X\rightarrow X$ is a map such that $f^{-1}[A]\in u$ for each $A\in u$, then $\{x\in X|f(x)=x\}\in u$. For a proof, see the book The Theory of Ultrafilters by Comfort and Negrepontis Thm 9.2 or Andreas Blass's dissertation p. 12.
A: To give a partial answer to Joel's challenge on the situation where the axiom of choice fails, we can have a case with a fixed point-free map. 
Suppose that $A$ is a $2$-amorphous set, that is an infinite set which cannot be written as the union of two disjoint infinite sets, but has a partition that almost all parts are pairs, and we can even assume that all parts are pairs. 
Note that the cofinite filter on $A$ is a free ultrafilter. Now fix a partition of $A$ into pairs, say $P$, and define the function $f(x)=y\iff\lbrace x,y\rbrace\in P$. Since this is a bijection cofinite sets are mapped to cofinite sets, but no point is a fixed point, as that would imply a singleton is in $P$.  
However this map has a period of two, so the main question is still open. 
