 Solution

Choose $s\in \mathbb{C}$ with $s^2=\lambda$. Since $\lambda \ne 0$, the invertible diagonal projective transformationfixes $[0:1:0]$. Writing its image as $[X:Y:Z]$, direct substitution givesThus it maps $C_{\lambda }$ onto $C_{1/\lambda }$. Either choice of square root works. We choose $s=i$ at $\lambda =-1$ and $s=e^{\pi i/6}$ at $\lambda =e^{\pi i/3}$ for the later order calculations.