Question about $S_\infty$ or $Aut(\mathbb{N})$

Question

I have been reading a little Kechris and other random Polish group books, and have come across a question I just can't wrap my mind around. Show that $S_\infty$ or $Aut(\mathbb{N})$ the set of all permutations, or bijections, on $\mathbb{N}$ has an element whose conjugacy class is dense in $S_\infty$. I have already shown that $S(\mathbb{N})_f$ the collections of permutations with finite support is dense and countable. I have a sneaky feeling it is supposed to use this fact, but have no clue how to do it. I have narrowed down to trying to show there is some $\pi \in S_\infty$ such that for all $\tau \in S_\infty$ and for all $F$ finite subsets of $\mathbb{N}$, there is $\sigma \in S_\infty$ such that $\sigma \pi \sigma^{-1} (f) = \tau(f)$ for all $f \in F$. Any helpful insight would be greatly appreciated.

Answer 1

Notice first that, if a permutation includes a cycle of length 17, then any sufficiently close approximation will have the same cycle. Therefore, the element you're looking for will have to include a 17-cycle. For the same reason, it will have to include infinitely many cycles of every finite length. I think (but I haven't checked carefully) that this is all you need, i.e., if a permutation has infinitely many cycles of every finite length, then its conjugacy class is dense. If I've messed up and this isn't true, then I recommend also having infinitely many infinite "cycles" in your permutation.