What classes of groups can arise as "symmetry groups of terms"? Let $\mathfrak{A}$ be an algebra (in the sense of universal algebra). To each term $t(x_1,...,x_n)$ in the language of $\mathfrak{A}$ in which each variable actually appears we can assign a group $G_\mathfrak{A}(t)\subseteq S_n$ consisting of all permutations of the variables which results in the same function: $$G_\mathfrak{A}(t)=\{\sigma\in S_n: \forall a_1,...,a_n\in\mathfrak{A}(t(a_1,...,a_n)=t(a_{\sigma(1)},...,a_{\sigma(n)})\}.$$
Now let $\mathbb{G}(\mathfrak{A})$ be the class of isomorphism types of groups of the form $G_\mathfrak{A}(t)$ for some term $t$. I'm curious which classes of groups can arise this way.
Originally I asked for a complete answer to this question, but in retrospect that was overly ambitious. To keep things reasonably answerable, let me restrict to the following natural candidates (playing fast and loose with isomorphism-type-vs.-group issues for simplicity) - I would love an answer, or even partial progress, to either question (or anything similar really, I'm profoundly stuck-in-the-weeds here):

Q1: Is there an $\mathfrak{A}$ such that $\mathbb{G}(\mathfrak{A})$ consists exactly of the finite $p$-groups for some prime $p$?

Certainly it's possible to get only $2$-groups - see e.g. this MSE answer of Eric Wofsey. However, getting exactly the $2$-groups, or $p$-groups for any fixed $p$, seems much harder.

Q2: Is there a $\mathfrak{B}$ such that $\mathbb{G}(\mathfrak{B})$ consists exactly of the finite abelian groups?

I really have no relevant information for this question, but it seems like a natural one to ask.
Note that Keith Kearnes' answer below addresses a variant of this question, in which variable appearance is replaced by variable dependence, and does not seem to immediately generalize to address this version.

Here are some example $\mathbb{G}(\mathfrak{A})$s (ignoring up-to-isomorphism details):

*

*If $\mathfrak{A}=(A;\star)$ where $\star:A^2\rightarrow A$ is a bijection, then $\mathbb{G}(\mathfrak{A})$ consists only of the trivial group.


*If $\mathfrak{A}=(\mathbb{N};\max)$, then $\mathbb{G}(\mathfrak{A})=\{S_n:n\in\mathbb{N}\}$.


*Although in the natural numbers with exponentiation alone there is essentially only one nontrivial equation, it turns out that every finite group shows up - see here.


*One of the few negative results I know is that the class of finite cyclic groups is not of the form $\mathbb{G}(\mathfrak{A})$ for any $\mathfrak{A}$. To see this, suppose $t(x_1,...,x_k)$ is a $\mathfrak{A}$-term with $G_\mathfrak{A}(t)\cong C_2$. Then the term $$s(x_{1,1},...,x_{k,k}):=t(t(x_{1,1},...,x_{1,k}), ..., t(x_{k,1},...,x_{k,k}))$$ has $G_\mathfrak{A}(s)$ having too many elements of order $2$ to be cyclic. In fact, tweaking this argument we get that $\mathbb{G}(\mathfrak{A})$ consists entirely of cyclic groups iff $\mathbb{G}(\mathfrak{A})$ consists only of trivial groups. However, this sort of idea doesn't seem to be useful for either Q1 or Q2 above.
 A: Let me edit this response in order to clarify what I am showing.

First, I will begin with an example: $t(x_1,x_2,x_3,x_4,x_5,x_6)
= ((((x_1x_2)x_3^{-1})x_3)x_4)x_4^{-1}$
is a group term. If you want to associate this term to a particular group, let it be the free group $F_6$ on $\{x_1,\ldots,x_6\}$. The term $t$ depends on $x_1$ and $x_2$.
The variables $x_3$ and $x_4$ occur,
but $t$ does not depend on them.
The variables $x_5$ and $x_6$ do not occur.
The permutations of indices that are symmetries of the original term $t$
are those from the full symmetric group $\textrm{Sym}(\{3,4,5,6\})$.
One may eliminate fictitious variables
by defining a term $s(x_1,x_2,x_3,x_4) = t(x_1,x_2,x_3,x_4,x_4,x_4)$.
The terms $s$ and $t$ are the same, all variables of $s$ occur in $s$, and $s$ does not depend
on its last two variables. The symmetries of $s$ are $\textrm{Sym}(\{3,4\})$. The term $r(x_1,x_2) = t(x_1,x_2,x_2,x_2,x_2,x_2)$ depends on all of its variables, which are the same variables that $t$ depends on, and $r$ acts the same way as $t$ with respect to those variables. The symmetry group of $r$ is trivial.
More generally, suppose that
$t(x_1,\ldots,x_i,\ldots,x_j,\ldots,x_n)$
depends on $x_1$--$x_i$, does not depend on $x_{i+1}$--$x_j$ although these variables occur,
and the variables $x_{j+1}$--$x_n$ do not occur.
The term $r(x_1,\ldots,x_i):=t(x_1,\ldots,x_i,x_i,\ldots,x_i)$
will be a term that depends on all variables and will
have symmetry group equal to some $G\leq S_i$.
The term $s(x_1,\ldots,x_j):=t(x_1,\ldots,x_i,\ldots,x_j,x_j,\ldots,x_j)$
will be a term equal to $t$, where all variables occur, and will
have the symmetry group $G\times S_{j-i}$.
The original term $t(x_1,\ldots,x_i,\ldots,x_j,\ldots,x_n)$
will have the symmetry group $G\times S_{n-i}$.
The symmetry groups of terms are determined, up to symmetric group factors, by the symmetry groups
of just those terms that depend on all variables. Moreover, the symmetric group factors can be determined if you know which variables of the term occur and which variables the term depends on.

There are various questions that could be asked, such as

* What are the possible symmetry groups of terms?
(Answer below: any concrete subgroup of $S_n$
is the symmetry group of an $n$-ary term that depends on all variables.)

* What are the possible classes of permutation groups
on finite sets which are the symmetry groups of those
terms of an algebra (i) which depend on all variables?
(ii) in which all variables occur? (iii) are arbitrary?
The question asked is (ii), but I will answer (i) here.
I am going to explain why, if $\mathcal K$ is any class of groups
whose members are subgroups of finite symmetric groups and $\mathcal K$ contains $S_1$, then there is an algebra $A$ such that the symmetry groups $G(t)$ of terms $t$ of $A$ that depend on all variables are exactly the groups in $\mathcal K$. (More precisely, if $G\in {\mathcal K}$ is a subgroup of $S_n$, then we will realize $G$ and all of its conjugates in $S_n$ as symmetry groups of terms of $A$ which depend on all variables.) We need to include $S_1$ in $\mathcal K$ because $t(x)=x$ has symmetry group $S_1$.

 
For a given $G\in\mathcal K$,
where $G\leq S_n$,
define an $n$-ary operation $f_G(x_1,\ldots,x_n)$ on the set
$\{a, b\}\cup \mathbb N^+
= \{a, b, 1, 2, 3, \ldots\}$ as follows:
$$
f_G(\bar{u}) = \begin{cases} a & \textrm{if $\bar{u} = 
(\sigma(1),\ldots,\sigma(n))$ for some $\sigma\in G$;}\\
b & \textrm{otherwise.}
 \end{cases}
$$
Let $A$ be the algebra on $\{a,b\}\cup \mathbb N^+$
whose operations are all operations
of the form $f_G$, $G\in\mathcal K$.

I claim that the following are true.

*

*Each $f_G$ depends on all of its variables.

*The symmetry group of $f_G(x_1,\ldots,x_n)$ is $G\;(\leq S_n)$.

*any term operation of $A$ that depends on all of its
variables is obtained from one of the $f_G$'s by
permuting the variables. Hence, its symmetry
group is conjugate to $G$ in $S_n$.
If $f_G(x_1,\ldots,x_n)$ is defined as above, and we identify two variables: $f_G(\underline{x_2},\underline{x_2},x_3,x_4,\ldots,x_n)$, then this operation is constant and its variable-set is $\{x_2,\ldots,x_n\}$.
Its symmetry group in Noah's sense is $S_{n-1}$. Similarly, the composite $f_G(f_G(x_1,\ldots,x_n),y_2,\ldots,y_n)$ is constant and its variable-set is $\{x_1,\ldots,x_n,y_2,\ldots,y_n\}$, so its symmetry group is $S_{2n-1}$. Using tricks like this one can obtain every finite symmetric group as a symmetry group of a term of $A$, as long as $A$ has at least one operation $f_G$ of arity greater than $1$.
The isomorphism types of Noah-symmetry groups turn out to be
exactly those in $\mathcal K\cup \{S_n\;|\;n=1, 2, \ldots\}$.
