$r_6$). For each $i$ let $W_i$ be the Weierstraß point of $C$ corresponding to the root $r_i$ of $f = g_1 g_2 g_3$. The kernel $H$ of the Richelot isogeny multiplication-by-$\sqrt{2}$ on the Jacobian $J$ of $C$ is the order-$4$ subgroup containing the divisor classes $[W_1 - W_2]$, $[W_3 - W_4]$, and $[W_5 - W_6]$.

Suppose there are distinct Galois-stable maximal isotropic subgroups $G$ and $G'$ of $({\operatorname{Jac}}C)[2]$ with $G' = (1+\sqrt{2})(G)$. Then clearly $G\neq H$, so $\#(G\cap H)$ is either $2$ or $1$. Suppose $\#(G\cap H)=2$. By renumbering the polynomials $g_i$ and by renumbering their roots, we may assume that $G$ is the order-$4$ subgroup $$G = \{ 0, [W_1 - W_2], [W_3 - W_5], [W_4 - W_6]\}.$$ Then $\sqrt{2}$ kills the first two elements of $G$ and sends the second two elements to $[W_1 - W_2]$, and it follows that $(1+\sqrt{2})(G) = G$, contradicting our assumption that $G$ and $G'$ are distinct.

So now we know that $G\cap H = \{0\}$. By renumbering the polynomials $g_i$ and by renumbering their roots, we may assume that $G$ is the order-$4$ subgroup $$G = \{ 0, [W_1 - W_5], [W_2 - W_4], [W_3 - W_6]\}.$$ It is not hard to 