= Solution
At $\lambda=-1$ and $s=i$, the chosen map is
$$
\sigma[x:y:z]=[-x:iy:z],\qquad
\sigma^2[x:y:z]=[x:-y:z],\qquad\sigma^4=1.
$$
The square is not the identity on the cubic, since a generic point has $y\ne0$. Thus $\boxed{\operatorname{ord}(\sigma)=4}$.
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 $1/\sqrt2<1$. The difference $\lambda_0-(m+ni)\omega$ belongs to $\Lambda$ and is shorter than $\omega$, so it must vanish. Hence
$$
\boxed{\Lambda=\omega(\mathbb Z+i\mathbb Z),\qquad
C_{-1}\cong\mathbb C/(\mathbb Z+i\mathbb Z).}
$$
The <lattice> is determined up to a nonzero complex scaling: it is square.
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