This is inspired by this question.
Does there exist an infinite finitely generated group having
(a) a unique
(b) finitely many
inclusion-minimal generating set(s) up to automorphisms?
♫
2.$ $ Which finite groups have a unique inclusion-minimal generating set up to automorphisms?
Here is a slight simplification of Geoff Robinson's nice proof that a finite group $G$ must be a $p$-group if its automorphism group acts transitively on the set of irredundant generating sets. First, as Robinson observed, the hypothesis implies that all members of every irredundant generating set have equal order, say $n$.
For every finite group, it is well known that the Frattini subgroup is the set of elements that are not part of an irredundant generating set, so in our case, the elements of $G$ outside of $\Phi(G)$ all have order $n$, and thus all elements of $G/\Phi(G)$ have order dividing $n$.
Next, we argue that $n$ is a prime power. Otherwise, let $g \in G - \Phi(G)$. Then $g$ has order $n$, so $g = xy$ for some elements $x$ and $y$, neither of which has order $n$. Then $x,y \in \Phi(G)$, so $g \in \Phi(G)$, a contradiction. Then every element of $G/\Phi(G)$ has $p$-power order, so $|G:\Phi(G)|$ is a power of $p$. It follows that $G = P\Phi(G)$, where $P$ is a Sylow $p$-subgroup of $G$, and thus $G = P$, as wanted.
As Benjamin Steinberg point out in his answer, this result is proved (probably more efficiently) in an Ohio State University undergraduate honors thesis (2012) by Paul Apisa
We claim in the finite case that this can only happen for $p$-groups, where $p$ is a prime : Suppose first that the finite group $G$ has a unique up to automorphism irredundant generating set of minimal cardinality, say $\{g_{1}, g_{2}, \ldots g_{n} \}.$ - this condition seems a priori weaker than that asked in the question. I claim that each $g_{i}$ has the same order. For if $g_{1}$ and $g_{2}$ have different orders, then $\{g_{1}, g_{1}g_{2}, g_{3},\ldots g_{n} \}$ and $\{g_{1}g_{2}, g_{2},g_{3}, \ldots g_{n} \}$ are also irredundant generating set of the same minimal cardinality, and the sum of the orders of the given generators is fixed by hypothesis, yielding a contradiction. Also, for each $i,$ we see that $\langle g_{i} \rangle$ must be maximal cyclic subgroup of $G,$ for if $\langle g_{i} \rangle$ is strictly contained in $\langle h_{i} \rangle , $ then $\{g_{1}, g_{2}, \ldots, g_{i-1}, h_{i},g_{i+1}, \ldots g_{n} \} $ is an irredundant generating set of the same minimal cardinality, but the sum of the order of its elements is larger than before. But now I claim in the context of the original question (in which we work from now on) that we may assume that each $g_{i}$ has (the same) prime power order. For if we could write $g_{1} = a_{1}b_{1} = b_{1}a_{1}$ with $a_{1},b_{1}$ of coprime orders, neither of which is $1,$ then by what we have already established, $\{a_{1},b_{1},g_{2},g_{3}, \ldots, g_{n} \}$ is an irredundant generating set for $G$ which clearly can't be obtained by applying an automorphism of $G$ to $\{g_{1},g_{2},\ldots, g_{n} \}.$ Now each $g_{i}$ has order $p^{k}$ for some fixed prime $p$ and positive (assuming $G$ is non-trivial!) integer $k$. Now every maximal subgroup of $G$ requires at least $n-1$ generators, by the minimality of $n.$ Furthermore, if there is a maximal subgroup $M$ of $G$ which requires $n-1$ generators, then every element of $G \backslash M$ has order $p^{k}.$ It follows in that case that $O^{p}(G) \leq M,$ and that $[G:M] = p$ (and $M \lhd G ).$ Now if every maximal subgroup of $G$ is normal, then $G$ is nilpotent, and hence is a $p$-group. Hence we may suppose that $G$ has a maximal subgroup $H$ which can be generated by $m \geq n, $ but no fewer, elements. Also, we may suppose that there is an element $y \in G \backslash H$ which does not have order $p^{k}.$ Let $\{h_{1},h_{2},\ldots h_{m} \}$ be an irredundant generating set for $H.$ We claim that $\{h_{1},h_{2},\ldots h_{m},y \} $ is an irredundant generating set for $G,$ which contradicts the transitivity of ${\rm Aut}(G)$ on irredundant generating sets. We certainly can't omit $y$ from the generating set. If we could omit one or more $h_{i},$ then we would have a minimal irredundant generating set obtaining an element not of order $p^{k},$ contrary to what we have already established. But now we have an irredundant generating set with $m+1 > n$ elements, still in contradiction to the transitivity assumption. Hence any finite group $G$ with this transitivity of ${\rm Aut}(G)$ on irredundant generating sets is a $p$-group for some prime $p.$
For question 2, let $p$ be a prime and let $\mathfrak A_p$ be the variety of elementary abelian $p$-groups. Let $\mathfrak A_p^m$ be the $m$-fold product of copies of this variety. It is locally finite, so has finite free objects. The relatively free group $G_n$ on $n$-generators should have the property that it has a unique inclusion-minimal generating set up to automorphism. The Frattini quotient is just $(\mathbb Z/p\mathbb Z)^n$ and so any inclusion-minimal generating set has $n$-elements. Mapping the basis to these $n$-elements gives an onto homomorphism which must be an automorphism by finiteness.
The same argument shows that in the profinite category, the free pro-$p$ group on $n$ generators has a unique inclusion-minimal generating set up to automorphism.
Added. Any finite $p$-group which is relatively free in some variety has this property by the same argument. In light of Geoff's answer perhaps these are the only ones?
Here is an honors thesis studying exactly this property for finite groups, which the author calls UMP, and all conclusions in the various answers are in the thesis. More examples are also given. A related paper by the same author is here
The free group $F_n$ satisfies the requirement (a) of the question 1. It is infinite finitely generated and its minimal generating set is unique up to automorphism.
To answer 1(a), if a finitely generated group $G$ has a unique inclusion minimal generating set $S$ then $G$ is finite. The reason is that if $g \not\in S$ then $g$ is a ``nongenerator'' in that it can be removed from any generating set (which must contain $S$) and the difference is still a generating set (because it contains $S$). But the set of nongenerators is a subgroup, the Frattini subgroup. The complement of the Frattini subgroup is therefore the finite set $S$. But that implies that the Frattini subgroup is finite, because for example the complement contains each left coset.
The same argument implies 1(b) using the union of the finite collection in place of just the one set $S$, although you need to precede that with an argument showing that an infinite generating set of a finitely generated group cannot be minimal, which is a well-known argument.
