Again, see How can I "see" that a map is birational? - but I formulate my hypothesis generically.

Let (the implicit surface) $F[x,y,z]$ be a polynomial with degree $2$ in all variables, i.e. $6$ overall. If $\forall_z F[x_1,y_1,z]=0$, we call $(x_1,y_1)$ a special pair for $z$. ($x_1=\infty$ etc. is allowed, but be careful.) Let's only consider all $F$ that have at least two special pairs for each variable. (Randall's equation in the above link is an example, e.g. $(0,1)$ and $(1,0)$ are special pairs for any variable.) For simplicity assume $F$ to be cyclic in the variables since I don't have an idea how my hypothesis must be formulated if not. Let

$G[x,y,x_1,y_1,x_2,y_2]=((x-x_1)/(x-x_2))\cdot((y-y_2)/(y-y_1))$

and similar for other variable pairs.

Consider the equation system

$G[x,y,x_1,y_1,x_2,y_2]=G[x',y',x_1,y_1,x_2,y_2]$

$G[y,z,y_1,z_1,y_2,z_2]=G[y',z',y,z,y_1,z_1,y_2,z_2]$

$G[z,x,z_1,x_1,z_2,x_2]=G[z',x',z_1,x_1,z_2,x_2]$

$F[x,y,z]=0$

It is fairly obvious that, given $(x,y,z)$ and solving for $(x',y',z')$, we get exactly two solutions, one being $(x,y,z)$. Call the other $X,Y,Z$. When is also $F[X,Y,Z]=0$?

Hypothesis $H$: Exactly if $(x_1,y_1)=(y_1,z_1)=(z_1,x_1),\dots$ (with $x_1!=x_2,\dots$) are indeed two different special pairs, $F[X,Y,Z]=0$.

This seems to work out for the Randall equation - I did not find yet other working examples than exactly those $G$ which can be built from special pairs, e.g. using $x\cdot(y-1)/(y+1)$ and cyclic for $G$. (Sidenote: Those aren't the only $1:1$ transformations, since I could throw $z$ into $G$ too, but those which are $1:1$ by construction.)

A brute force approach with MATHEMATICA was hopeless (and also would only "prove" it for one $F$ at a time unless I use coefficients for $F$ too). Thus maybe some math is in order. (Looks like a case of projective geometry for me...) Can you prove $H$ (maybe even in a generalized form not assuming $F$ is cyclic)?

Again, see How can I "see" that a map is birational? — but I formulate my hypothesis generically.

Let (the implicit surface) $F[x,y,z]$ be a polynomial with degree $2$ in all variables, i.e. $6$ overall. If $\forall_z F[x_1,y_1,z]=0$, we call $(x_1,y_1)$ a special pair for $z$. ($x_1=\infty$ etc. is allowed, but be careful.) Let's only consider all $F$ that have at least two special pairs for each variable. (Randall's equation in the above link is an example, e.g. $(0,1)$ and $(1,0)$ are special pairs for any variable.) For simplicity assume $F$ to be cyclic in the variables since I don't have an idea how my hypothesis must be formulated if not. Let $$ G[x,y,x_1,y_1,x_2,y_2]=((x-x_1)/(x-x_2))\cdot((y-y_2)/(y-y_1)) $$ and similarly for other variable pairs.

Consider the equation system \begin{gather*} G[x,y,x_1,y_1,x_2,y_2]=G[x',y',x_1,y_1,x_2,y_2] \\ G[y,z,y_1,z_1,y_2,z_2]=G[y',z',y,z,y_1,z_1,y_2,z_2] \\ G[z,x,z_1,x_1,z_2,x_2]=G[z',x',z_1,x_1,z_2,x_2] \\ F[x,y,z]=0 \end{gather*}

It is fairly obvious that, given $(x,y,z)$ and solving for $(x',y',z')$, we get exactly two solutions, one being $(x,y,z)$. Call the other $(X,Y,Z)$. When is also $F[X,Y,Z]=0$?

Hypothesis H: Exactly if $(x_1,y_1)=(y_1,z_1)=(z_1,x_1),\dotsc$ (with $x_1\ne x_2,\dotsc$) are indeed two different special pairs, $F[X,Y,Z]=0$.

This seems to work out for the Randall equation — I did not find yet other working examples than exactly those $G$ which can be built from special pairs, e.g. using $x\cdot(y-1)/(y+1)$ and cyclic for $G$. (Sidenote: Those aren't the only 1:1 transformations, since I could throw $z$ into $G$ too, but those which are 1:1 by construction.)

A brute force approach with Mathematica was hopeless (and also would only "prove" it for one $F$ at a time unless I use coefficients for $F$ too). Thus maybe some math is in order. (Looks like a case of projective geometry for me….) Can you prove H (maybe even in a generalized form not assuming $F$ is cyclic)?