This turned up with the Euler Brick.
It is well known that solving $(p^2-1)(q^2-1)(r^2-1)-8pqr=0$ in rational numbers gives an Euler brick (a quader with rational sides and space diagonals). The symmetry group is huge, one triple is equivalent to $384-1$ other. One can reparametrize very much, my favorite variables are $Z=(p+1/p)(q+1/q)(r+1/r)$ and $U=(too long)$. Now I can do the following: If I hold $Z$ constant (all equivalent triples give $Z$ or $-Z$), for a smaller set of the triples also $(1+(U+Z)^2)*U^2$ is an invariant (which is essentially the square of the space diagonal). Likewise I can hold $U+Z$ and $(1+U*Z+U^2)/U$ constant (same) or $U$ and $(-1-4*U^2-4*U^4+6*U*Z- 4*U^3*Z-Z^2+U^2*Z^2)/(1+(U+Z)^2)$ (an interesting involution that sends the trivial triple $(0,1,c)$ to the Euler solution if you take the limit properly). All solutions after solving after $(p,q,r)$ again are rational.
Now it is near lying trying to combine the invariants, so for example if $I$ is the invariant, $I=I',Z=Z'$ works as well as $I=I',U=U'$ (say), so triples with the same $Z$ as well as triples with the same $U$ have the same $I$.
Is there a general method for invariant combining and what $I$ will result?