 Solution

At $\lambda =-1$ and $s=i$, the chosen map isThe square is not the identity on the cubic, since a generic point has $y\ne 0$. Thus $\boxed{\mathrm{ord}(\sigma )=4}$.

Choose the cubic's point $[0:1:0]$ as the origin in its complex torus description. Since $\sigma$ fixes it, its torus form is $z+\Lambda \mapsto az+\Lambda$, with $a\Lambda =\Lambda$. Exact order four implies $a=\pm i$; $a^2=-1$. Therefore $\Lambda$ is stable under multiplication by $i$.

To identify its shape without an unproved uniqueness assertion, choose a shortest nonzero period $\omega$. For any ${\lambda }_0\in \Lambda$, approximate ${\lambda }_0/\omega$ by a Gaussian integer $m+ni$ within distance $1/\sqrt{2}<1$. The difference ${\lambda }_0-(m+ni)\omega$ belongs to $\Lambda$ and is shorter than $\omega$, so it must vanish. HenceThe lattice is determined up to a nonzero complex scaling: it is square.