I did some computations with the integers up to 400000 and I got the following conclusions:

1) The following are cycles with more than one element (i.e. non-fixed points)

- [32,80]=[2⁵, 2⁴·5]
- [864,2160]=[2⁵·3³, 2⁴·3³·5]
- [708588,2598156,787320]=[2²·3¹¹, 2²·3¹⁰·11, 2³·3⁹·5]
- [5832,17496,61236,20412]=[2³·3⁶, 2³·3⁷, 2²·3⁷·7, 2²·3⁶·7]

2) All the integers up to 400000 end up falling in a fix point or one of the previous cycles (or maybe another 2-cycle that I have missed) except for $2^{16}$, $2^{17}$, $2^{16}\cdot 3$, $2^{16}\cdot 5$, $2^{18}$, $2^{17}\cdot 3$, $2^{12}\cdot 3^4$, $2^4\cdot 3^9$, $2^9\cdot 3^6$.

These exceptional values end up reaching some number that my computer cannot handle. As an example the first six exceptional values (all of those with the exponent of 2 being bigger or equal to 16) lead to $2^{18}\cdot 3^5$, which after 10 more steps or so computed by hand keeps increasing and doesn't look like falling into a cycle soon. This agrees with Felipe's comment.

3) I found all the above using bad programming: I computed the actual numbers instead of keeping track of primes and exponents as one does by hand. Reprogramming everything using this idea should give much better results.

**Addendum 1:** Studying the case $2^{16}$ I could find some more 2-cycles, such as

- $[2^{63}\cdot 3^{13}\cdot 7\cdot 31,2^{62}\cdot 3^{14}\cdot 7\cdot 13]$ ($2^{16}$ reaches this 2-cycle after 33 steps).
- $[2^{279}\cdot 3^{61}\cdot 31\cdot 139,2^{278}\cdot 3^{62}\cdot 31\cdot 61]$
- $[2^{15}\cdot 3^{33}\cdot 7\cdot 17,2^{14}\cdot 3^{34}\cdot 5\cdot 11]$

The left term of all of them is of the form $2^{2p+1}\cdot 3^{2q-1}\cdot p\cdot q$ with $(2p+1)(2q-1)=9\cdot c$ where $p,q$ are primes different from $2,3$ and $c$ is a square-free integer coprime with $2,3$. More generally we have the following.

*Let $p,q,r,s$ be pairwise distinct primes such that $(pr+1)(ps-1)=q^2\cdot c$ where $c$ is a square-free integer coprime with $p,q$. Then $[p^{pr}\cdot q^{ps}\cdot c,p^{pr+1}\cdot q^{ps-1}\cdot r\cdot s]$ is a $2$-cycle of $f$.*

N.B.: If $r,s\neq 2$, the hypothesis force either $p=2$ or $q=2$. So a much easier result is:

*Let $p>2,3$ be a prime such that $2p=3c+1$, where $c$ is a square-free integer coprime with $3$. Then $[2^3\cdot 3^{2p-1}\cdot p,2^2\cdot 3^{2p}\cdot c]$ is a $2$-cycle of $f$.*

E.g. $p=11,17,29,47,53$ do the trick.

**Addendum 2:** I have already coded everything properly (please tell me if you are interested in the code). At the moment I'm looking for cycles as long as possible. If found out some 5-cycles and 6-cycles: for example 2²·3⁴⁷·5⁶ is the beginning of a 5-cycle and 2⁶⁷·3¹⁰⁵·5⁵·53·67 is the beginning of a 6-cycle. Of course one wonders if there are cycles of arbitrary length and how to find them. It would be nice to find a proposition extending the one I gave in Addendum 1 to $f^k$ for arbitrary $k$.

**Addendum 3:** I got similar results for $k=4$:

*Let $p>2,3$ be a prime such that $6p+1$ is a square-free integer. Then $[2^3\cdot 3^{6p}\cdot p, 2^3\cdot 3^{6p+1}\cdot p, 2^2\cdot 3^{6p+1}\cdot (6p+1),2^2\cdot 3^{6p}\cdot(6p+1)]$ is a $4$-cycle of $f$.*

All the primes between 5 and 97 satisfy the conditions of this proposition except for 29 and 79.

*Let $p>2,3$ be a prime such that $p-1=6c$ for a square-free integer $c$. Then $[2^2\cdot 3^p\cdot p, 2^2\cdot 3^{p-1}\cdot p, 2^3\cdot 3^{p-1}\cdot c, 2^3\cdot 3^p\cdot c]$ is a $4$-cycle of $f$.*

Some primes which satisfy the conditions are 31, 43, 67 and 79.