Possible symmetry groups of power terms

Question

Previously asked and bountied at MSE:

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Let $\mathfrak{E}=(\mathbb{N};\mathit{exp})$ be the algebra in the sense of universal algebra consisting of the natural numbers with just exponentiation. To each term $t(x_1,...,x_n)$ in which each variable $x_i$ ($1\le i\le n$) actually appears$^1$ we can assign the group $$E_t=\{\sigma\in S_n:\forall a_1,...,a_n\in\mathbb{N}[t(a_1,...,a_n)=t(a_{\sigma(1)},...,a_{\sigma(n)})]\}.$$ For example, allowing standard notational conveniences we have $E_{x^y}$ is trivial but $E_{(x^y)^z}\cong S_2$.

I'm curious which groups arise, up to isomorphism, as $E_t$s (in the language of this earlier question of mine, I'm asking for a description of $\mathbb{G}(\mathfrak{E})$). The above trick is the only useful thing I can think of, and in a sense is in fact all there is, but it already gives rise to some complexity: for example, at a glance the term $$[[((a^b)^c)^d]^{[((p^q)^r)^s]}]^{[((w^x)^y)^z]}$$ yields a semidirect product of $(S_3)^3$ and $S_2$, but we can then "carve out" some of that group by reusing the same variable multiple times. Intuitively I suspect that each $E_t$ can be built up from full permutation groups via semidirect products + [something else rather simple], but it seems potentially messy. There are many specific groups which seem (to me) to be plausible counterexample candidates, including the $A_n$s and $C_n$s for "large enough" values of $n$, but I haven't had any luck figuring out the situation with even such fairly simple low-complexity groups.

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$^1$The answer to this specific question would not change if we allowed terms in which some variables don't appear; however, for general structures $\mathfrak{A}$ this restriction can be impactful (e.g. if we take $\mathfrak{A}$ to be an algebra consisting of a single bijection from the square of the underlying set to itself), so I've included it here for consistency.

Answer 1

Based on an observation by MSE user Pilcrow, it seems I've been overcomplicating this:

For simplicity, let "$[x_1,x_2,...,x_k]$" be shorthand for the right-associating exponent term $$x_1^{(x_2^{(...^{x_k})})}.$$ Then for each $k\in\mathbb{N}$ and each subgroup $G$ of $S_k$, we can consider the term $$t_G:=w^{\prod_{\sigma\in G}[x_{\sigma(1)},x_{\sigma(2)}...,x_{\sigma(k)}]}$$ (allowing the obvious abuse of notation for brevity), with $w,x_1,x_2,...,x_k$ distinct variables.

Since $w$ obviously can't be swapped with any of the $x_i$s there is a canonical embedding $i:E_{t_G}\rightarrow S_k$, and it's not hard (if a bit tedious) to show that in fact we have $i[E_{t_G}]=G$. So every finite group occurs as the symmetry group of some exponentiation-only term. For example, $C_4$ is represented by $w^{[x_1,x_2,x_3,x_4]\cdot[x_2,x_3,x_4,x_1]\cdot[x_3,x_4,x_1,x_2]\cdot[x_4,x_1,x_2,x_3]}$, or a bit less abbreviatedly by $$w^{[x_1^{(x_2^{(x_3^{x_4})})}]\cdot[x_2^{(x_3^{(x_4^{x_1})})}]\cdot[x_3^{(x_4^{(x_1^{x_2})})}]\cdot[x_4^{(x_1^{(x_2^{x_3})})}]}.$$