Let $G$ be a finite group of order $n$ and let $\Delta$ be its generating set. I'll say that $\Delta$ *generates $G$ symmetrically* if for every permutation $\pi$ of $\Delta$ there exists $f:G\rightarrow G$ an automorphism of $G$ such that $f\restriction\Delta=\pi$.


How large can $\Delta$ be with respect to $n$? Specifically what's the asymptotic behaviour? Is there a class of groups (along with their symmetric generating sets) of unbounded order such that $n$ is polynomially bounded by $|\Delta|$. Has any other research been done on these generating sets?



How large can a symmetric generating set of a finite group be?
Cherry-picking the answer from YCor's and Colin Reid's comments:


If $G$ has order $n$, then $|\Delta| \leq \log\_2 n + 1$. This bound is sharp in the sense that it is attained for a sequence of groups of unbounded order (namely elementary abelian 2-groups).


Proof:


Let $G$ be a finite group and let $\Delta$ be a generating set with the property described in the OP (i.e. any permutation of $\Delta$ induces an automorphism of $G$).


Let $k$ be the cardinality of $\Delta$ and let $\Delta = \{g\_1,\dots,g\_k\}$. (Note this fixes a total order on $\Delta$.)


For any $j=1,\dots,k$, let $\Delta\_j := \{g\_1,\dots,g\_j\}$ and let $G\_j$ be the subgroup generated by $\Delta\_j$. For $j\leq k-2$, $G\_j$ must generate a proper subgroup of $G\_{j+1}$, because $G$ has automorphisms that fix $G\_j$ pointwise but do not fix $G\_{j+1}$ pointwise (namely, those induced by any permutation of $\Delta$ that fixes $\Delta\_j$ pointwise but moves $g\_{j+1}$; note that for $j\leq k-2$, these exist).


As a consequence, $[G\_{j+1}:G\_j]\geq 2$ for $j\leq k-2$.


Also note $|G\_1|\geq 2$ since $g\_1$ cannot be the identity as there exist automorphisms of $G$ that do not fix it (as long as $|\Delta|\geq 2$).


Thus, by induction, $|G\_j|\geq 2^j$ for $j$ up to $k-1= (k-2)+1$. In particular, $G\supset G\_{k-1}$ must have order at least $2^{k-1}$.


This yields the bound given above.


This bound is attained for elementary abelian 2-groups of order $n=2^{k-1}$, for $k\geq 3$: let $\Delta$ consist of a basis (of $k-1$ elements) plus the product of the basis elements. (The condition $k\geq 3$ guarantees the product is distinct from any of the basis elements.) This choice of generators satisfies the condition $\prod g\_i = 1$ inside the group, thus the basis elements $g\_1,\dots,g\_{k-1}$ can be sent to any $k-1$ of the $k$ elements of $\Delta$, inducing an automorphism, and the last element of $\Delta$ will automatically end up in the right place.


