I would say no, as such a map would not be induced by a single map $F_n \to F_{n-1}$. Indeed, any map $F_n \to F_{n-1}$ is, by a Theorem 3.3, p. 132, in Magnus-Karrass-Solitar's _Combinatorial Group Theory_, essentially just the map $\varphi : F_n \to F_{n-1}$ got by killing a coordinate in $F_n$ (that is, by selecting a different free generating set for $F_n$, we may suppose the map is precisely $\varphi$ on these new generators). Thus, one can use the example in my [previous answer][1] to give an IA automorphism whose image is not an isomorphism.

Thus, we cannot hope that anything very close to the Braid group case works here. However, it may be possible to show that for any IA automorphism, $f$, there is some map $F_n \to F_{n-1}$, depending on $f$, that in turn sends $f$ to an automorphism, but I'm not sure if I'd call this natural.


  [1]: https://mathoverflow.net/questions/163971/property-of-ia-automorphisms-of-free-groups/163989#163989