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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 permutation 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 $s(x_1,\ldots,x_i):=t(x_1,\ldots,x_i,x_i,\ldots,x_i)$ will be a term that depends on all variables and will have symmetry group equal to some $G\leq S_i$. The term $r(x_1,\ldots,x_j):=t(x_1,\ldots,x_i,\ldots,x_j,x_j,\ldots,x_j)$ 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.

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.

I will explain why any collection of isomorphism types of subgroups of symmetric groups can be realized as $\mathbb G(A)$ for some algebra $A$.

But first, let me address this phrase from the problem statement: in which each variable actually appears. This condition does not restrict. If an algebra $A$ has a term $t(x,y,z)$ where only $x$ and $y$ `actually appear', then one can expand $A$ by adding a new fundamental operation $f$ interpreted so that $f(x,y,z) := t(x,y,z)$ on $A$. This has no effect on which functions are the interpretations of $n$-ary terms for $A$, but it does have an effect on whether variables `actually appear' in the term: each of $x$, $y$, and $z$ actually appears in $f$, while $f$ has the same symmetry group as $t$.

Instead of

Now, back to

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 $S_n$.
  

This is enough to show that

Edit. Noah didn't like me changing his question, so let me say what the above construction contributes

If

Let me edit this response in order to clarify what I am showing.

First

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 $s(x_1,\ldots,x_i):=t(x_1,\ldots,x_i,x_i,\ldots,x_i)$ will be a term that depends on all variables and will have symmetry group equal to some $G\leq S_i$. The term $r(x_1,\ldots,x_j):=t(x_1,\ldots,x_i,\ldots,x_j,x_j,\ldots,x_j)$ 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

I will explain why any collection of isomorphism types of subgroups of symmetric groups can be realized as $\mathbb G(A)$ for some algebra $A$.

But first, let me address this phrase from the problem statement: in which each variable actually appears. This condition does not restrict. If an algebra $A$ has a term $t(x,y,z)$ where only $x$ and $y$ `actually appear', then one can expand $A$ by adding a new fundamental operation $f$ intepreted so that $f(x,y,z) := t(x,y,z)$ on $A$. This has no effect on which functions are the interpretations of $n$-ary terms for $A$, but it does have an effect on whether variables `actually appear' in the term: each of $x$, $y$, and $z$ actually appears in $f$, while $f$ has the same symmetry group as $t$.

Instead of the condition on term operations in which each variable actually appears I shall work with the condition on term operations which depend on all variables.

Now, back to the problem. I claimed above that any collection of isomorphism types of subgroups of symmetric groups can be realized as $\mathbb G(A)$ for some algebra $A$. This is almost true. You must include the isomorphism type of the $1$-element group, since it is the symmetry group of $t(x)=x$. So let $\mathcal K$ be any class of subgroups of finite symmetric groups which includes the $1$-element group. 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 $S_n$.
  

This is enough to show that $\mathbb G(A)$ consists of exactly the isomorphism types of groups in $\mathcal K$.

Edit. Noah didn't like me changing his question, so let me say what the above construction contributes to the problem in its original form.

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 $\mathcal K\cup \{S_n\;|\;n=1, 2, \ldots\}$.

I will explain why any collection of isomorphism types of subgroups of symmetric groups can be realized as $\mathbb G(A)$ for some algebra $A$.

But first, let me address this phrase from the problem statement: in which each variable actually appears. This condition does not restrict. If an algebra $A$ has a term $t(x,y,z)$ where only $x$ and $y$ `actually appear', then one can expand $A$ by adding a new fundamental operation $f$ interpreted so that $f(x,y,z) := t(x,y,z)$ on $A$. This has no effect on which functions are the interpretations of $n$-ary terms for $A$, but it does have an effect on whether variables `actually appear' in the term: each of $x$, $y$, and $z$ actually appears in $f$, while $f$ has the same symmetry group as $t$.

Instead of the condition on term operations in which each variable actually appears I shall work with the condition on term operations which depend on all variables.

Now, back to the problem. I claimed above that any collection of isomorphism types of subgroups of symmetric groups can be realized as $\mathbb G(A)$ for some algebra $A$. This is almost true. You must include the isomorphism type of the $1$-element group, since it is the symmetry group of $t(x)=x$. So let $\mathcal K$ be any class of subgroups of finite symmetric groups which includes the $1$-element group. 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 $S_n$.
  

This is enough to show that $\mathbb G(A)$ consists of exactly the isomorphism types of groups in $\mathcal K$.

Edit. Noah didn't like me changing his question, so let me say what the above construction contributes to the problem in its original form.

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 $\mathcal K\cup \{S_n\;|\;n=1, 2, \ldots\}$.