Let $\Lambda(\lambda) := \left\{\lambda, 1 - \lambda, \frac{1}{\lambda}, \frac{1}{1-\lambda}, \frac{\lambda - 1}{\lambda}, \frac{\lambda}{\lambda - 1} \right\}$, and consider the $j$-function
$$j(\lambda) = 256\frac{(\lambda ^2 - \lambda + 1)^3}{\lambda ^2 (\lambda - 1)^2}.$$
I'm trying to prove that $\Lambda(\lambda) = \Lambda(\lambda ')$ iff $j(\lambda) = j(\lambda ')$.
That the values of the $j$-function coincide if the sets are the same is fairly obvious by a straightforward calculation, but I'm stuck trying to prove the converse.
I'd welcome any hints.
$j(\lambda)$ is a rational function of degree $6$, so for a generic $z$ there a $6$ $\lambda$'s such that $j(\lambda)=z$, and for the critical values there are less than $6$ $\lambda$'s. If $\lambda$ is such that $\Lambda(\lambda)$ has $6$ elements and if $j(\lambda)=j(\lambda')$, it follows that $\lambda'\in\Lambda(\lambda)$. If $\Lambda(\lambda)$ has less than $6$ elements, you can check it manually (there are only two possible values of $j$ in this case) [there is a better explanation, but perhaps this one will do :) ]