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Not a full answer. Just describing what happens at the level of rational functions as an answer, because it does not fit into a comment. Invariant theory has a lot to say about this and related problems.

Let $E=K(x_1,x_2,\ldots,x_n)$ by the field of rational functions in $n$ independent indeterminates. Let $F=K(s_1,s_2,\ldots,s_n)$ be the subfield of symmetric rational functions. It is well known that $E/F$ is a Galois extension with Galois group $G\cong S_n$ acting by permuting the indeterminates.

Consider the element $$ u=\frac{x_1}{x_2}+\frac{x_2}{x_3}+\cdots+\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}. $$ It is easy to see that $\sigma(u)=u$ if and only if $\sigma$ is a power of the $n$-cycle $\alpha=(123\ldots n)$. By basic Galois theory this implies that $F(u)$ is the fixed field $E^H$ of $H=\langle\alpha\rangle$. Alternatively, you can show that the orbit of $u$ under $G$ has $(n-1)!$ elements.

We can bring all the fractions in $u$ together as follows $$ u=\frac{\sum_{i=1}^nx_i^2\prod_{j,j

eq i, j

ot\equiv i+1\pmod n}x_j}{x_1x_2\ldots x_n}. $$ The denominator is in $E$. It may be tempting to conjecture that elementary symmetric polynomials together with the numerator might generate the ring of polynomial invariants. Unfortunately I don't have any intuition about this. It need not be that simple.

In the case of finite groups generated by reflections (= finite Coxeter groups), there is a beautiful description of the polynomial invariants due to Chevalley. See for example Chapter 3 of J.E. Humphreys, Reflection Groups and Coxeter Groups, Cambridge studies in advanced mathematics #29. I suspect the answer to the present question is also known, but I don't have the time to search for it.