Under what conditions there exists a topology making a bijection into a homeomorphism? Let $X$ be a set. Assume we have a bijection $f:X \rightarrow X$ which is not the identity. 
I want to find necessary or sufficient conditions on $f$ and $X$, that should imply existence of a topology on $X$ making $f$ a homeomorphism.
Of course there are the two trivial topologies that come to mind (Every set is open, and nothing is open except for $\emptyset,X$).
So I wonder if there is something interesting to say, after we exclude the above trivialities ($f\neq Id$, non-trivial topologies).
 A: To start with, what follows is little more than a dictionary translation of the condition that such a topology should exist: for each $S \subset X$, consider the class of sets $$\mathcal A_S =\{f^i(S) \mid i \in \mathbb Z\},$$ where for $i>0$, $f^i(S)$ is the image of $S$ under the $i^\text{th}$ iterate of $f$, and for $i<0$, $f^i(S)$ is the $(-i)^\text{th}$ preimage of $S$ under $f$.
If $f$ is a homeomorphism and $S$ is to be open, then the family of sets $\mathcal A_S$ must also be open. So it is natural to consider the topology $\tau_S$ generated by $\mathcal A_S$, which has basis given by all finite intersections of elements of $\mathcal A_S$.
It's not difficult to show that $f$ is a homeomorphism under the topology $\tau_S$, so the question simply becomes: does there exist an $S \subset X$, $S \neq \emptyset, X$, such that $\tau_S \neq \mathcal P(X)$?
With that in mind, it's easy to give some finite examples for which the answer is no: the permutation $(12 \cdots p)$ on $\{1, 2, \ldots, p\}$ for $p$ a prime number has no such topology.
If $f$ is given by the $n$-cycle $(12 \cdots n)$, and $n=ab$ is not prime, then the sets $\{i, i+a, \ldots, i+a(b-1)\}$ for $1 \le i <a$ partition $\{1, 2, \ldots, n\}$ into subsets that are permuted by $f$. So there is a nontrivial topology in this case.
For general $f$ on a finite set $X$, you simply look at the cycle decomposition of $f$ and use the above cases.
For infinite sets, you always have the finite complement topology, since $f$ is assumed to be a bijection.
So it turns out this question on its own isn't very interesting. Of course, there are some natural follow-up questions you could ask that could potentially make it interesting in the infinite case; for instance, what if you want the topology to be Hausdorff?
