I have been studying iterated polynomials, specifically let $P(x)=x^2+c$ and consider the equation $P(P(P(x)))=x$. After dividing out the solutions of $P(x)=x$, we have six solutions, which obey the following equations: \begin{equation} \begin{aligned} x_{2a}=P(x_{1a})\\ x_{3a}=P(x_{2a})\\ x_{1a}=P(x_{3a})\\ x_{2b}=P(x_{1b})\\ x_{3b}=P(x_{2b})\\ x_{1b}=P(x_{3b}) \end{aligned} \end{equation} These equations are symmetric under $G_0$, a subgroup of the permutation group. The generators of $G_0$ are two rotations: \begin{equation} x_{1a}\mapsto x_{2a} \mapsto x_{3a} \mapsto x_{1a} \\ x_{1b}\mapsto x_{2b} \mapsto x_{3b} \mapsto x_{1b} \end{equation} and the interchange \begin{equation} \begin{aligned} x_{1a} \mapsto x_{1b} \mapsto x_{1a}\\ x_{2a} \mapsto x_{2b} \mapsto x_{2a}\\ x_{3a} \mapsto x_{3b} \mapsto x_{3a} \end{aligned} \end{equation}
My question is what are the generators of polynomials invariant under $G_0$. There are many polynomials invariant under $G_0$ but not the permutation group $P_6$. These include \begin{equation} \begin{aligned} s1=(x_{1a}+x_{2a}+x_{3a})(x_{1b}+x_{2b}+x_{3b})\\ s2=x_{1a} x_{2a}+x_{2a} x_{3a}+x_{3a}x_{1a}+x_{1b} x_{2b}+x_{2b} x_{3b}+x_{3b}x_{1b}\\ s3=x_{1a} x_{2a}^2+x_{2a} x_{3a}^2+x_{3a}x_{1a}^2+x_{1b} x_{2b}^2+x_{2b} x_{3b}^2+x_{3b}x_{1b}^2 \end{aligned} \end{equation} Note that $s1$ and $s2$ are invariant under single interchanges such as $x_{1a} \mapsto x_{2a} \mapsto x{1a}$, which is not in $G_0$. $s3$ is only invariant under $G_0$.
Again, to be clear, what is a set of polynomials (which will likely include $s1, s2$, and $s3$)such that any polynomial invariant under $G_0$ can be expressed algebraically as an expression in $s1, s2, s3, ...$? Is this set finite? Related question; can $s1, s2, s3, etc.$ be expressed as rational functions of c? Remember c? $P(x)= x^2+c$.