Finite order automorphisms of surface groups preserving a representation

Denote by $\Gamma_g = \pi_1(\Sigma_g)$ the fundamental group of the surface of genus $g \ge 2$, and equip it with its usual presentation $$\Gamma_g = \langle a_1, b_1, \dots, a_g, b_g \mid [a_1, b_1] \dots [a_g, b_g] \rangle$$ For $2 \le n \le g$, we denote by $F_n = \langle A_1, \dots, A_n \rangle$ the free group of rank $n$. We consider the morphism $$\rho : \begin{array}[t]{rcl} \Gamma_g & \to & F_n \\ a_i, i \le n & \mapsto & A_i \\ a_i, i > n & \mapsto & 1 \\ b_i & \mapsto & 1 \end{array}$$ Is there a non-trivial finite order automorphism $\varphi$ of $\Gamma_g$ such that $\rho \circ \varphi = \rho$ ?

Thanks for any pointers!

• What about $S_{g-n}$ acting by permutations on $\{a_i:i>n\}$ and $\{b_i:i>n\}$? – arctic tern Aug 26 '16 at 21:33
• Unfortunately, I don't think this defines a morphism of $\Gamma_g$, since the word relation isn't preserved by these permutations. – George P Aug 26 '16 at 22:41
• Ignoring $n$ for the moment, shouldn't $S_g$ preserve the word relations? For instance if $g=2$, then the map defined by ($a_1\leftrightarrow a_2$ and $b_1\leftrightarrow b_2$) should preserve the truth of the set of relations $[a_1,b_1],[a_2,b_2]$. – arctic tern Aug 27 '16 at 2:25
• Not all the permutations : only the $g$-cycles (corresponding to moving commutators around the word) and the products of permutations $(a_1 \leftrightarrow b_g) (b_1 \leftrightarrow a_g) \dots$ (corresponding to taking the inverse of the word). – George P Aug 27 '16 at 10:25
• Any map which takes $b_i$ to the identity has the effect $[a_i,b_i] = a_i b_i a_i^{-1} b_i^{-1} \to a_i 1 a_i^{-1} 1 = 1$. So any map which takes every $b_i$ to the identity has the effect $[a_1,b_1]...[a_g,b_g] \to 1$. – Lee Mosher Aug 27 '16 at 13:58