 Solution

Let $(r_0,r_1,r_2)$ be any ordering of the three distinct numbers $0,1,\lambda$. Set $h=r_1-r_0$ and choose $s\ne 0$ with $s^2=h^3$. The projective transformationhas the required matrix form. As $x=hX+r_{0}Z$ and $y=sY$, substitution givesThe choice $s^2=h^3$ is exactly what cancels the coefficient introduced by the change in $x$.

The six orderings give the Legendre parameter orbit:Every entry is different from $0,1$ when $\lambda \ne 0,1$.

There are no further possibilities for a matrix of the stated form. Its determinant is $acd$, so $a,c,d$ must be nonzero. On $z=1$ it induces the affine change $u\mapsto (a/d)u+b/d$ and scales $v=y/z$ by $c/d$. It preserves the line $y=0$, whose three intersections with the plane cubic have $u$-coordinates $0,1,\lambda$. Their images must have coordinates $0,1,\tilde{\lambda }$. Choosing the two roots sent to $0,1$ determines the affine change and leaves exactly the six orderings above.

For general $\lambda$ the six values are distinct. They coincide in some special cases: $\lambda \in \{-1,2,1/2\}$ gives three values, and ${\lambda }^2-\lambda +1=0$ gives two. In particular, for the value obtained in part (c), the two possible values are $(1\pm \mathrm{i}\sqrt{3})/2$. This does not contradict the paper's qualification that there are six for general $\lambda$.