When does a clone on a two-element set have almost abelian symmetry groups?

Question

Say that a clone (in the sense of universal algebra) $\mathfrak{C}$ has almost abelian symmetry groups (= aasg) iff for each function $f(x_1,...,x_n)\in\mathfrak{C}$ there is an abelian subgroup $A\subseteq S_n$ and a set $X\subseteq \{1,...,n\}$ such that $$\langle A\cup \mathsf{Fix}(X)\rangle=\{\sigma: f=\lambda x_1,...,x_n. f(x_{\sigma(1)},...,x_{\sigma(n)})\}$$ (where "$\mathsf{Fix}(X)$" denotes the subgroup of $S_n$ consisting of permutations fixing $X$ pointwise and "$\langle U\rangle$" denotes the subgroup of $S_n$ generated by $U$). Basically, $\mathfrak{C}$ has aasg if every function in $\mathfrak{C}$ has an abelian symmetry group once we ignore "dummy variable" issues.

(Note that we're looking at the symmetry groups of the elements of the clone, not the symmetry group of the clone - in any sense - itself.)

Coming from this old question of mine, I'm interested in understanding aasg-ness. It seems natural to start with the case where we look at clones on a $2$-element set, since things are already nontrivial but well-understood there:

Question 1: Which clones on a $2$-element set have aasg?

Note that Post's lattice consists of finitely many elements together with eight infinite families, so - since aasg-ness is inherited by subclones - there are only a few different "shapes" that the set of aasg clones could have. In particular, the set of aasg clones on a two-element set is (in the appropriate sense) computable (thanks to Emil Jerabek for pointing this out!).

A follow-up question is whether, like general clones, things "blow up" once we get to three elements:

Question 2: How many aasg clones are there on $\{1,2,3\}$? If there are only countably many, what is the least finite $n$ (if any) such that there are continuum-many aasg clones on $\{1,...,n\}$?

Answer 1

Question 1: Is the set of clones on $\{1,2\}$ which have aasg computable?

The answer is Yes. The aasg clones are just the essentially unary clones on $\{1,2\}$, and there are six of them.

Easy direction: Assume that $\mathfrak C$ is essentially unary and $f(x_1,\ldots,x_n)\in \mathfrak C$ does not depend on its first $n-1$ variables. The symmetry group of $f$ is $\textrm{Sym}(\{1,\ldots,n-1\})$ if $f$ depends on its last variable and is $\textrm{Sym}(\{1,\ldots,n\})$ otherwise. Both cases satisfy the aasg property.

Less easy direction: Assume that $\mathfrak C$ is a clone on $\{0,1\}$ that is not essentially unary. [I changed my underlying set to $\{0,1\}$ so that I can write operations as Boolean expressions.] I will argue that (i) $\mathfrak C$ contains a totally symmetric operation of arity $\geq 3$ that depends on all variables, and that (ii) a clone with a totally symmetric operation of arity $\geq 3$ that depends on all variables fails the aasg property.

For (i), just examine Post's lattice. Every clone that is not essentially unary contains one of the Boolean operations

(a) $x\wedge y\wedge z$ (ternary conjunction),
(b) $x\vee y\vee z$ (ternary disjunction),
(c) $x \oplus y \oplus z$ (ternary addition modulo $2$),
(d) $(x\wedge y)\vee (x\wedge z)\vee (y\wedge z)$ (ternary majority).

For (ii), assume that $\mathfrak C$ contains a totally symmetric operation $f(x_1,\ldots,x_m)$ which depends on all variables and that $m\geq 3$. For $n\geq 3$, the symmetry group of $g(x_1,\ldots,x_m,x_{m+1},\ldots,x_{m+n}):=f(x_1,\ldots,x_m)$ is $S_m\times S_n$, and both factors are nonabelian.

Claim. The symmetry group of $g$, which is $S_m\times S_n\leq S_{m+n}$, cannot have the form $\langle A\cup \textrm{Fix}(X)\rangle$ for any abelian subgroup $A\subseteq S_{m+n}$ and any $X\subseteq \{1,\ldots,m+n\}$.

Assume otherwise. Note that $\textrm{Fix}(X)$ is generated by the transpositions it contains, and that this set of transpositions is `connected' in the sense that $(i\;j), (k\;l)\in\textrm{Fix}(X)$ implies $(j\;k)\in \textrm{Fix}(X)$. The only connected subsets of transpositions in $S_m\times S_n$ lie in $S_m\times \{1\}$ or $\{1\}\times S_n$, so $\textrm{Fix}(X)$ lies in either $S_m\times \{1\}$ or $\{1\}\times S_n$. This is enough to see that if $N$ is the normal subgroup $S_m\times S_n$ that is generated by $\textrm{Fix}(X)$, then $(S_m\times S_n)/N$ is nonabelian, which can't happen if $S_m\times S_n=\langle A\cup \textrm{Fix}(X)\rangle$.