Let $P=P_n$ be the pure braid group on $n$ strands and $F=F_n$ the free group on $n$ generators. I'm interested in a nice description of the action of $P$ on the derived subgroup $F'$ which somehow would'nt involve $F$.

More precisely, there is well known action of $P$ on $F$ which has the very particular property of being basis conjugating, i.e. it send every generator of $F$ to a conjugate of itself. In fact, in the following feel free to replace $P$ by the group of all basis conjugating automorphism (which has a somewhat simpler presentation).

This property implies that $P$ acts trivially on the abelianization $F/F'$ (though it is much stronger). It means that $$\forall p \in P, f\in F,\ (p\cdot f)f^{-1}\in F'\ \ (*)$$ where $\cdot$ is the action.

Let $\gamma_k F$ be the $k$th term of the lower central series, so that $\gamma_1F=F$, $\gamma_2F=F'$ and $$\gamma_{k+1}F=[\gamma_kF,F].$$

Then it's a general fact that $(*)$ implies that $\gamma_kP$ acts trivially on $\gamma_lF$ modulo $\gamma_{k+l}F$.

In particular, we have $$\forall p \in P', f\in F',\ (p\cdot f)f^{-1}\in \gamma_4 F$$

but I suspect that it actually belongs to a much smaller group. In fact I was hoping that it belonged to the commutator subgroup of $F'$ $$[F',F']\subset \gamma_4F$$

i.e. that the action of $P'$ on $F'$ would have the same property than the action of $P$ on $F$, but that doesn't seem to be true (I would be happy to be proven wrong here !). So my question can be asked in two different direction:

Is there a nice description/are there nice properties of the action of $P$, or any higher term $\gamma_kP$, on $F'$ given directly in term of $F'$ ?

One can do some pretty explicit computation but it didn't give me any insight yet. In the other but somewhat more concrete direction:

What is the subgroup of $P$ that acts on $F'/F''$ trivially ? Is it non trivial ? Does it contains $\gamma_kP$ for some $k$ ?

Some googling told me that this last question might be related to the Gassner representation of $P_n$, and the so-called magnus representation of the Torelli group of a surface, but I haven't found any answer to this question yet.