A marked graph is a pair $(G,τ)$ where $G$ is a graph and $τ: R_n \to G$ is a homotopy equivalence . $R_n$ is the Rose with n petals which is isomorphic to $F_n$ ,the free group with $n$ generators.
A homotopy equivalence $σ: G_2 \to G_1 $ is called Out-inverse to a homotopy equivalence $τ:G_1 \to G_2$ if for any vertices $u_1 \in V(G_1)$ and $u_2 \in V(G_2)$ the maps $σ \circ τ $ and $τ \circ σ $ induce the identity outer Automorphism of the groups $π_1(G_1,u_1)$ and $π_1(G_2,u_2)$ respectively.
Let $(G,τ)$ be a marked graph and let $σ : G \to R_n $ a homotopy equivalence Out-inverse to $τ$. Then every homotopy equivalence $f: G \to G$ determines the outer automorphism $(σ \circ f \circ t)_\circledast$ of the group $π_1(R_n,\ast)= F_n$. This outer automorphism does not depend on the choice of $σ$.
My main question is: why this outer automorphism does not depend on choice of $σ$ ?
This statement is the same to say that that the homotopy equivalnce $σ \circ f \circ t: R_n \to R_n$ determines the outer automorphism $(σ \circ f \circ t)_\circledast$ of the group $π_1(R_n,\ast)$ ?