Assuming we're talking about the same map and there's just a $n \longmapsto n+1$ mix-up, this question reduces to studying $Inn(F_n)$ intersected with the image of $B_n \to Aut(F_n)$.

An inner automorphism of $F_n$ is a conjugation automorphism, so we're looking at braids that act on $F_n$ by conjugation. Let the generators of $F_n$ be denoted $x_1, \cdots, x_n$ and consider the product $x_1x_2\cdots x_n$. The image of the braid group fixes this element (the image of $B_n \to Aut(F_n)$ is precisely the subgroup of $Aut(F_n)$ that fixes this element). But if $c x_1x_2 \cdots x_n = x_1x_2 \cdots x_n c$, $c$ must be a power of $x_1x_2\cdots x_n$ since otherwise $F_n$ wouldn't be a free group.