= Solution
Choose $s\in\mathbb C$ with $s^2=\lambda$. Since $\lambda\ne0$, the invertible diagonal <projective transformation>
$$
\boxed{\sigma_\lambda[x:y:z]=[x/\lambda:y/(\lambda s):z]}
$$
fixes $[0:1:0]$. Writing its image as $[X:Y:Z]$, direct substitution gives
$$
Y^2Z=\lambda^{-3}y^2z
=X(X-Z)(X-\lambda^{-1}Z).
$$
Thus 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.
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