^1 (a)= \varphi (a, {\mathrm{id}}, \dots, {\mathrm{id}})$, respectively. Then, the induced group homomorphisms $\gamma _*$ and $\gamma ^1 _*$ on homotopy $\pi_q (\operatorname{Aut}_H (F_H)) \to \pi _q (\operatorname{Aut}_H (\ast _k F_H)) $ satisfy the relation $\gamma _*=k \gamma^1 _*$.

Let $\gamma ^i: \operatorname{Aut}_H (F_H)\to \operatorname{Aut}_H (\ast _k F_H )$ be the map defined by $$\gamma ^i (a)=\varphi ({\mathrm{id}}, \dots, a, \dots, {\mathrm{id}})$$ where $a$ is on the $i$-th coordinate. We have $\gamma=\gamma ^1 \gamma ^2 \cdots \gamma ^k$ under the product induced by the product in the monoid $A$. Since the group operation on $\pi _q (\operatorname{Aut}_H (\ast _k F_H))$ coming from the monoid structure on $\operatorname{Aut}_H (\ast _k F_H)$ coincides with the usual group structure on homotopy groups, we have $\gamma _*= \gamma ^1 _* +\cdots + \gamma ^k _*$. So, to complete the proof, it is enough to show that $\gamma ^i$ and $\gamma ^j$ are homotopic for every $i,j \in \{ 1, \dots, k\}$. Since $F_H$ is $H$-homotopy equivalent to ${\mathbb S}(V_H)$, it is enough to prove this for ${\mathbb S}(V_H)$. Note that in this case we have $\gamma ^i=T(i,j) \gamma ^j$ where $T(i,j): V_H^{\oplus k} \to V_H^{\oplus k}$ is a linear transformation which swaps the $j$-th 