Elliptic curve with 6 order automorphism

Question

I have a question about automorphism of elliptic curve. Let $\langle1,\zeta \rangle$ with $\zeta^6\,=\,1$ be a lattice. The automorphism should save the lattice corresponding to the curve. But why the group $\mathbb{Z}_6$ saves this lattice? The parallelogram $0,1,\zeta+1,\zeta$ comes to $0,\zeta,2\zeta-1,\zeta-1$ by muliplying by $\zeta$ and it does not belong to the lattice, I suppose. What's wrong with it? Thanks!

Answer 1

We don't need the automorphism to preserve the fundamental parallelogram. It just needs to map lattice points to lattice points (and be invertible). So for example, $2\zeta -1$ on the image parallelogram is a linear combination of $\zeta$ and $1$ on the first.