Let $S\_n$ act on the set of boolean functions of size $n$ in the following way:


If $f$ is a boolean function and $\alpha \in S\_n$, then $g=\alpha f$ and $g(x)=f(\alpha(x))$ where $x$ is boolean vector of size $n$;


Let $p$ be the number of orbits. Since each orbit can have no more than $n!$ elements it is obvious that $p\ge\frac{2^{2^n}}{n!}$.


I am interested in an upper bound of $p$. Do you know any related results?



Symmetric group acting on the set of boolean functions
The number of orbits of a two-cycle $\sigma$ acting on subsets of
$\{1,\dots,n\}$ is $3\cdot 2^{n-2}$. If follows that the number of orbits of
$\sigma$ acting on boolean functions is $2^{2^{n-1}}\cdot 3^{2^{n-2}}$.
By the Cauchy-Frobenius lemma
(a.k.a. Burnside's lemma), it is then easy to see that
 $$ p =\frac{1}{n!}\left( 2^{2^n}+\binom n2 2^{2^{n-1}}\cdot 3^{2^{n-2}}
 + o\left(2^{2^{n-1}}\cdot 3^{2^{n-2}}\right)\right). $$
Thus for instance for all $\varepsilon>0$ and sufficiently large $n$
(depending on $\varepsilon$), we have
 $$ p<\frac{1}{n!}\left( 2^{2^n}+(1+\varepsilon)\binom n2 2^{2^{n-1}}\cdot 3^{2^{n-2}} \right). $$


