I've got a polynomial transformation on $\mathbb{R}^6$, and I have a conjecture about it, but I'm having a hard time proving it. The transformation looks like this:
$ u:= abcde + abc + abe + ade + cde + a + c + e\\ v:= bcdef + bcd + bcf + bef + def + b + d + f\\ w:= cdefa + cde + cda + cfa + efa + c + e + a\\ x:= defab + def + deb + dab + fab + d + f + b\\ y:= efabc + efa + efc + ebc + abc + e + a + c\\ z:= fabcd + fab + fad + fcd + bcd + f + b + d,$
so as you can see, all six formulas are the same, up to a cyclic permutation of the variables. (This arose in the study of periodic continued fractions, but I don't think that really matters right now.)
Now, it's clear that if the variables $a,b,c,d,e,f$ satisfy the relations $\{a=c=e, b=d=f\}$, this would imply the corresponding relations $\{u=w=y, v=x=z\}$. Is the converse true?
I think it should be, but every way that I try to prove it leads me into horrible thickets of algebra with no way out. Does anyone see a nice way to argue this from symmetry? If the claim is not true, then what's a good strategy for producing a counterexample? I'm only interested in cases where all variables are strictly positive, otherwise we could get a straightforward counterexample using $0$'s.
A related question: Do the relations $\{u=x,v=y,w=z\}$ imply the relations $\{a=d,b=e,c=f\}$?
With $a=c=f=0$, your equations $u=w=y, v=x=z$ become $bde=0$. So take $b=0$ and $d \ne 0$.
EDIT: With the additional constraint that all variables are strictly positive, the answer is yes. Using Maple, I took a "plex" Groebner basis of the ideal generated by the polynomials $u-w,w-y,v-x,x-z$, and the first element factored as $f e^2 d (d-f) (c d e f+c d+c f+e f) (c d e f+c d+c f+e f+2)$. If all variables are strictly positive, the only way this can be $0$ is $d=f$. Substituting $d=f$ into the fourth basis element and factoring gives $ce{f}^{3} \left( ef+2 \right) \left( cf+1 \right) \left( c-e \right) \left( cef+2\,c+e \right)$ so we get $c=e$. Then substituting $d=f$ and $c=e$ into the 15'th basis element gives $bf{e}^{2} \left( -f+b \right)$ so $b=f$. And finally, substituting $d=f$, $c=e$, $b=f$ into the $21$'st basis element gives $e{f}^{2} \left( -e+a \right)$ so $a=e$.
ANOTHER EDIT: In principle it might be possible to attack your second question the same way, but so far (after a couple of hours) Maple 17 is still working on a Groebner basis for this one.
FURTHER EDIT: It did produce a basis. The first basis element is $\left( def+d-e+f \right) \left( c-f \right) \left( cde+c-d+e \right) \left( cdef+cd+cf+ef+2 \right) \left( cdef+cd+cf+ef \right)$, so besides $c=f$ we could have $e = (d+f)/(1-df)$ with $df < 1$ or $e = (d-c)/(1+cd)$ with $c < d$.
After some fiddling, I found a solution $\left\{ a=1/2,b=2,c=3/4,d=1,e=2,f=1/3 \right\}$ to $\{u=x,v=y,w=z\}$.
AND MORE: $\{a=3,b=1/2,d=3,f=1,e=(3-c)/(3c+1)\}$ for $0 < c < 3$.