*Previously [asked and bountied at MSE](https://math.stackexchange.com/questions/4338046/what-are-the-symmetry-groups-of-exponentiation-only-terms):*




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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](https://mathoverflow.net/questions/412563/what-term-symmetry-groups-can-an-algebra-admit), 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](https://math.stackexchange.com/a/2080920/28111) 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.



Possible symmetry groups of power terms
Based on [an observation by MSE user Pilcrow](https://math.stackexchange.com/questions/4358618/do-cyclic-groups-appear-as-symmetry-groups-of-exponentiation-only-terms#comment9100944_4358618), 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 $$w^{[x\_1^{(x\_2^{(x\_3^{x\_4})})}]\cdot[x\_2^{(x\_3^{(x\_4^{x\_1})})}]\cdot[x\_3^{(x\_4^{(x\_1^{x\_2})})}]\cdot[x\_4^{(x\_1^{(x\_2^{x\_3})})}]}.$$


