= Solution
On the parameter, the two transformations are $S(\lambda)=1/\lambda$ and $T(\lambda)=1-\lambda$. Their compositions yield
$$
\lambda,\quad S\lambda=1/\lambda,\quad T\lambda=1-\lambda,
\quad TS\lambda=1-1/\lambda,\quad ST\lambda=1/(1-\lambda),
\quad STS\lambda=\lambda/(\lambda-1).
$$
Every step lifts to an invertible <projective transformation> of the corresponding cubic <curves>. Therefore \b[all six parameter values give isomorphic <curves>]. Some values coincide for special $\lambda$; this does not affect the conclusion.
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