\nabla_{p_2}S_h(p,0,x)$ and $G_p(x) = \nabla_{p_1}S_h(0,p,x)$. Then $F_p$ and $G_p$ are formal diffeomorphisms and their inverses are given by $$F^{-1}_p(x) = \nabla_{p_2}S_h(-p,p,x),\quad G^{-1}_p(x) = \nabla_{p_1}S_h(p,-p,x).$$

Denote $\overline F_p(x) = \nabla_{p_2}S_h(-p,p,x)$ and $\overline G_p(x) = \nabla_{p_1}S_h(p,-p,x)$. Differentiating the SGA equation w.r.t. $p_1$, putting $p_1=p$, $p_2=-p$, $p_3=p$, we get that $\overline G_p\circ G_p = id$. Putting $p_1=0$, $p_2=p$, $p_3=-p$, we get $G_p\circ \overline G_p = id$. Thus $\overline G_p = G^{-1}_p$. Similarly differentiating the SGA equation w.r.t. $p_3$, putting $p_1=p$, $p_2=-p$, $p_3=p$, we get that $\overline F_p\circ F_p = id$. Putting $p_1=p$, $p_2=-p$, $p_3=0$, we get $F_p\circ \overline F_p = id$. Thus $\overline F_p = F^{-1}_p$.

\[relationformdiffeo\] Denote $F_p(x) = \nabla_{p_2}S_h(p,0,x)$ and $G_p(x) = \nabla_{p_1}S_h(0,p,x)$. Then we have the relation $$F_p = G_{-p}.$$

Notice that it is equivalent to prove that $F_p = G_{-p}$ or $F^{-1}_p = G^{-1}_{-p}$. We prove the second identity. For each $n\geq 1$ we have the decomposition $$S_n(p_1,p_2,x) = S_n^1(p_1,p_2,x)+  S_n^2(p_1,p_2,x)+\dots+ S_n^n(p_1,p_2,x)$$ where $S_n^i$ is the part of $S_n$ which is homogeneous of degree $i$ in the first argument. Now we have that $$S_n^i(-p,p,x)= (-1)^iS_n^i(p,p,x) = 0$$ because of naturality of the generating function. This implies that $S(-p,p,x) = 0$. If we differentiate this equation with respect to p 