How large can a symmetric generating set of a finite group be?

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?

Is there any difference between these two solutions for finding generating function?

Consider the following question.

Find the generating function for the number of integer solutions to the equation $ c_{1}+c_{2}+c_{3}+c_{4}=20$ where $ -3\leq c_{1}, -3 \leq c_{2}, -5\leq c_{3}\leq 5, $ and $ 0 \leq c_{4}$ .

  • The first solution: $ $ (c_{1} + 3)+(c_{2}+3)+(c_{3}+5)+c_{4}=20 + 3 + 3 + 5 \Rightarrow c_{1}’+c_{2}’+c_{3}’+c_{4}’=31 $ $ where $ 0\leq c_{1}’, 0 \leq c_{2}’, 0\leq c_{3}’\leq 10, $ and $ 0 \leq c_{4}’$ . The generation function is $ f(x) = (x^{0}+x^{1}+…)^{3}(x^{0}+x^{1}+…+x^{10})$ , and the number of integer solutions is the coefficient of $ x^{31}$ in $ f(x)$ .

  • The second solution (My solution): $ $ g(x) = (x^{-3}+x^{-2}+…+x^{3})^{2}(x^{-5}+x^{-4}+…+x^{5})(x^{0}+x^{1}+…)$ $ , and the number of integer solutions is the coefficient of $ x^{20}$ in $ g(x)$ .

My question: In the solution manual book, the presented solution is the first solution, but I think that the second solution is correct, and it is simpler than the first solution. Is my solution correct? What is the difference between these two solutions?