I posted another question here with description for the specific case math.stackexchange.com/questions/3156893/…

In general we can assume that they are known and part of the input. But for the specific case (the action I described at the end of question) all that functions are defined and maybe there are some tricks that work in that specific case. Any help will be useful.

@YCor We can assume that there is a function that gets element of $g\in G$ and element of $s\in S$ and returns the result when $g$ acts on $s$. By better algorithm I mean algorithm that calls that function less.

Yes it is the group generated by $n$-length cycle permutation. It has $n$ powers of $(1,2,\ldots,n)$

I thought to get a covering using idea in this question. math.stackexchange.com/questions/2441363/…

One question. $\binom n2 2^{2^{n-1}}\cdot 3^{2^{n-2}}$ is number of orbits when a single two-cycle acts on the set of boolean functions, right? By Cauchy-Frobenius lemma we should add number of functions that a group element leaves unchanged. I am missing something?