= Solution
Take $\lambda=e^{\pi i/3}$ and the chosen $s=e^{\pi i/6}$. Then $\lambda s=i$, so $i/(\lambda s)=1$, and $-1/\lambda=\lambda^2$. In affine coordinates the chosen lift becomes
$$
\upsilon(x,y)=(1+\lambda^2x,y).
$$
Since $\lambda^2$ is a primitive cube root of unity, $1+\lambda^2+\lambda^4=0$. Three iterations fix $x$ and $y$, whereas one does not. Thus
$$
\boxed{\operatorname{ord}(\upsilon)=3\quad\text{for the lift chosen above}.}
$$
The other square-root choice replaces this map by $(x,y)\mapsto(1+\lambda^2x,-y)$. Its cube is $(x,y)\mapsto(x,-y)$, so that lift has order six. This sign dependence is why a lift was specified explicitly.
Both fix the point at infinity. In the <complex torus> model their linear multipliers have order three or six, so the <lattice> is stable under a primitive cube root $\zeta=e^{2\pi i/3}$ and hence under $e^{\pi i/3}=1+\zeta$. Choose a shortest nonzero period $\omega$. The triangular <lattice> $\mathbb Z+\zeta\mathbb Z$ has covering radius $1/\sqrt3<1$: its equilateral triangles of side one each have circumradius $1/\sqrt3$. Approximating any period divided by $\omega$ by this <lattice> leaves a strictly shorter period, which must be zero. Therefore
$$
\boxed{\Lambda=\omega(\mathbb Z+e^{\pi i/3}\mathbb Z).}
$$
Thus the cubic has the hexagonal <lattice> shape, independently of which of the two lifts is chosen.
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