I don't know much about free groups (excepted the very basics), and the following question may be trivial, although it isn't to me. Let $F$ be a free group with $n$ generators $x_1,\dots,x_n$. Consider the 'augmentation' map $a:F \rightarrow \mathbb{Z}$ that sends $x_i$ to $1$ for $i=1,\dots,n$, and let $A = \ker f$. So $A$ is a free group as a subgroup of the free groups $F$, of infinite rank I presume. Is it easy to describe a free family of generators of $A$? This would surely allow to answer my question, which is:

Can we describe the action of $\mathbb{Z}$ on $A^{ab}$ ?

Here, $A^{ab} = A /D(A)$ is the abelianization of $A$ (with $D(A)$ its derived subgroup), and the action in question is the one given by the short exacts sequence: $$1 \rightarrow A/D(A) \rightarrow F/D(A) \rightarrow \mathbb{Z} \rightarrow 1$$ The group $F/D(A)$ is the metabelian group of the title. Of course, describing the action is essentially equivalent to describing the group $F/D(A)$, since the extension splits.

(The question feels elementary to me, but at the same time I feel helpless to solve it, because I don't know how to recognize when a family of elements in a free group is free.)

Now the truth is that the real question I need is when $F$ is a free *pro-$p$* group
with $n$ generators instead of a free group, $a$ is the *continuous* map from $F$ to $\mathbb{Z}_p$ that sends the generators to $1$, and $D(A)$ is the closed derived subgroup. But I believe (perhaps naively) that the solution of the discrete problems will easily give a solution of its pro-$p$ analog, and that the discrete problem is more natural.

This question in turn comes from my trying to understand the structure of the maximal metabelian quotient of some pro-$p$ Galois group of number fields with prescribed ramification. In some cases such a group is the $p$-adic $F/D(A)$ considered in this question.