Fixed points of polynomial ring homomorphism 
$S=\mathbb R[x+y+z, xy+yz+zx, xyz]$ is the ring of the symmetric polynomials in $\mathbb R[x,y,z]$. Let $\psi\colon S \to R[x,y,z]$ be a ring homomorphism such that
  \begin{align}
x &\mapsto -x,\\
y &\mapsto z-x, \\
z &\mapsto y-x.
\end{align}
  I'd like to know how to find the fixed points of $\psi$, that is the polynomials $p$ such $\psi(p)=p$, or equivalently such that $p(x,y,z)=p(-x,z-x,y-x)$ (apart from the constant polynomials).

We have 
\begin{align}
a = x+y+z &\mapsto -3x+y+z = A\\
b= xy+yz+zx &\mapsto 3  x^2 - 2 xy - 2xz + y z = B\\
c= xyz &\mapsto -x^3 + x^2 y + x^2 z - x y z = C .
\end{align}
So far I've found that $C= - c + x^2 (A-2x)$ and $B+Ax= - b + 2 yz$, but I cannot easily get rid of those $x,y,z$, and actually I'm just blindly trying to find some relations and hopefully some fixed points. Even if I found any of those polynomials, I still wouldn't know if there aren't others... Any help to untangle this mess would be very appreciated, as well as some reference on the topic of invariant polynomials :)
 A: You want to find polynomials that are invariant by both $\psi$, and the natural action of $S_3$ on $\Bbb R[x,y,z]$, so it's natural to look at the subgroup $G$ of $Aut(\Bbb R[x,y,z])$ generated by $\psi$ and $S_3$.
It turns out that $G$ is isomorphic to $S_4$. 
Since our copy of $S_3$ in $G$ looks just like a copy of $S_3$ in $S_4$ (a normalizer of an element in the set on which $S_4$ acts), $G$ is the group of permutations of $G/S_3$. A natural polynomial invariant by $S_3$ is $T = x+y+z$, its other images by $G$ are then $X = -3x+y+z, Y = x-3y+z$, and $Z = x+y-3z$.
Now then it is easy to check that $G$ is the group of permutations of $\{X;Y;Z;T\}$, because the action of $S_3$ on $x,y,z$ is translated into the same action on $X,Y,Z$ ; and $\psi(X,Y,Z,T) = (T,Z,Y,X)$
Now let $X,Y,Z,T$ be indeterminates and let $\phi : \Bbb R[X,Y,Z,T] \to \Bbb R[x,y,z]$ be defined by $\phi(X) = -3x+y+z$ etc. $G$ acts on both rings and its action is compatible with $\phi$ ($\phi$ is a map of $G$-modules).
$\phi$ has a right-inverse $\sigma :\Bbb R[x,y,z] \to \Bbb R[X,Y,Z,T]$ defined for example by $\sigma(x) = (T-X)/4$ etc, which shows that $\phi$ is surjective.
In fact one can show that $\sigma$ is also a map of $G$-modules (this is pretty clear for the $S_3$ part of $G$ so you only need to check that $\psi$ commutes with $\sigma$)
This shows that $\Bbb R[X,Y,Z,T]$ is the direct sum of $\ker \phi$ and $\sigma(\Bbb R[x,y,z])$ (as $G$-modules), and that all the invariants in $\Bbb R[x,y,z]$ come from invariants in $\Bbb R[X,Y,Z,T]$.
But it is well-known that $\Bbb R[X,Y,Z,T]^G = \Bbb R[E_1,E_2,E_3,E_4]$ where the $E_i$ are the elementary symmetric polynomials in $X,Y,Z,T$.
$\phi(E_1) = 0$, (in fact $\ker \phi = (E_1)$) ;  but you get
$\phi(E_2) = -6(x^2+y^2+z^2)+4(xy+yz+zx)$
$\phi(E_3) = -8(x^3+y^3+z^3)+8(x^2y+x^2z+xy^2+xz^2+y^2z+yz^2)-16xyz$
$\phi(E_4) = -3(x^4+y^4+z^4)+4(x^3y+x^3z+xy^3+xz^3+y^3z+yz^3) +14(x^2y^2+y^2z^2+z^2x^2) -20xyz(x+y+z)$
