This came up with the Euler brick.

Let $(p,q,r)$ be a Randall triple, i.e. $(p^2-1)(q^2-1)(r^2-1)=8pqr\ [eq.1]$. There are tons of maps that map a triple to another: obvious ($p\rightarrow -p,q\rightarrow 1/q,r\rightarrow r)$, still easy ($p\rightarrow (p+1)/(p-1),q\rightarrow (q+1)/(q-1),r\rightarrow (r+1)/(r-1))$ and outrageous ($p\rightarrow (\dots+p^2q^2r^2)/(\dots+p^2q^2r^2)$, likewise $q,r$ - I found three such maps).

Thus let $F[p,q,r]$ be a broken rational function of $p,q,r$ where the maximum exponent of each variable in numerator and denominator is $2$ (so overall $6/6$), and $p'=F[p,q,r],q'=F[q,r,p],r'=F[r,p,q]$. This makes up for a whopping $54$ free coefficients, so it is hopeless to check if $p',q',r'$ fulfils Eq.1 by plugging it in, do a resultant with Eq.1 to eliminate $r$ and set the coefficients (estimated: up to $p^{100}$) of $p,q$ to zero.

My maps are not only birational, but even involutions. Are there indicators for "F is birational" or "F is even an involution" that help eliminating a bucketload of coefficients first?

The following might help to massacre a few coefficients beforehand: $T=(0,1,s)$ is a valid triple for any $s$, so $F$ must map $T$ either to the other $23$ triples that can be obtained by using the above trivial maps on $T$, or $F$ must give $(p',q',r')=(0/0,0/0,0/0)$ (or another indeterminate) which evaluates to a meaningful limit which is a parametric solution (e.g. the Saunderson one).