Symmetries of truth functions

Question

The symmetry group of a given $n$-ary truth function $f:\{0,1\}^n\to\{0,1\}$ is the permutation group consisting of all the $f$-preserving permutations of the $n$ input variable names: $\{\sigma\in S_n:f^\sigma=f\}$, where $f^\sigma(x_1,\dots,x_n):=f(x_{\sigma(1)},\dots,x_{\sigma(n)})$. For example, the truth function $f(x_1,x_2,x_3)=x_1\land(x_2\lor x_3)$ has symmetry group generated by the transposition $(23)$.

Does every subgroup of $S_n$ occur in this way?

Note that the permutation structure, not just the group structure, matters here. For example, $\langle(1234)\rangle$ and $\langle(1234)(56)\rangle$ are isomorphic as groups, but $\langle(1234)\rangle$ ought to be the symmetry group of something like

$$f(w,x,y,z)=g(w\to x,x\to y,y\to z,z\to w)$$

for some highly-symmetric $g$ (and possibly a higher-arity asymmetric operation in place of "$\to$"), whereas I can't think of a truth function with symmetry group generated by $(1234)(56)$.

Answer 1

Edit: The previous argument that was here was incorrect, but I don't think so. Boolean functions $f : \{ 0, 1 \}^n \to \{ 0, 1 \}$ are in bijection with subsets of $\{ 0, 1 \}^n$ (namely the subset $f^{-1}(1)$), or equivalently with sets of subsets of an $n$-element set $[n]$, and this bijection preserves symmetries; that is, the symmetries of $f$ are the symmetries of the corresponding set of subsets of $[n]$. Write $F_k \subset {[n] \choose k}$ for the set of subsets in $f^{-1}(1)$ of size $k$. Then the condition that $\sigma \in S_n$ is a symmetry of $f$ is equivalent to the condition that $\sigma F_k = F_k$ (setwise; my mistake previously was confusing this with pointwise) for all $k$.

The choice of collections of subsets $F_k$ amounts to a choice of (undirected) hypergraph structure on $[n]$, so the question is equivalent to asking whether every subgroup of $S_n$ appears as the automorphism group of a hypergraph structure on $[n]$.

I claim that $A_n$ is not the automorphism group of a hypergraph structure for $n \ge 3$. It suffices to observe that $A_n$ acts transitively on the set ${[n] \choose k}$ of subsets of size $k$, for all $1 \le k \le n-1$. (For $1 \le k \le n-2$ this follows from but is weaker than the statement that $A_n$ acts $k$-transitively, which involves ordered tuples. For $k = n-1$ it follows from $1$-transitivity.) It follows that if a hypergraph has automorphism group containing $A_n$ then for each $k$ it contains either all $k$-edges (subsets of size $k$) or none of them, so its automorphism group is $S_n$.