Finite linear group invariant polynomials separate orbits

The orbits are disjoint: if $gx=hy$, then $y=h^{-1}gx$ and their full orbits agree, contrary to distinctness. Let $T=O_1\cup O_2$ and fix finite linear coordinates $x_1,\dots ,x_r$. For each $b\in O_2$ and $a\in T\setminus \{b\}$, choose the least coordinate index $j(a,b)$ with $x_j(b-a)\mathrm{\ne }0$. Such an index exists because coordinates separate distinct vectors. Define ${\mathrm{\ell }}_{a,b}(x)=(x_{j(a,b)}(x)-x_{j(a,b)}(a))\mathrm{/}(x_{j(a,b)}(b)-x_{j(a,b)}(a))$. Its denominator is nonzero, it is zero at $a$, and it is one at $b$.

For each $b\in O_2$ put ${\delta }_b={\prod }_{a\in T\setminus \{b\}}{\mathrm{\ell }}_{a,b}$, with empty product one. At $b$ every factor is one; at any other $a\in T$ the corresponding factor vanishes. Thus $q={\sum }_{b\in O_2}{\delta }_b$ is zero on $O_1$ and one on $O_2$. Set $p=\mathcal{R}(q)$ using F1. It is invariant. For $x\in O_j$ and every $g\in G$, $g^{-1}x$ stays in $O_j$, so every term $q(g^{-1}x)$ has the required constant value there. Averaging preserves that value, proving the assertion.

Distinct orbits are nonempty, so $T$ has at least two points and no empty coordinate family is needed: in dimension zero only one orbit exists and the hypothesis cannot hold. If either orbit is a singleton, the same products work, including the orbit of the zero vector. Stabilizers never alter interpolation coefficients because $T$ contains distinct points; Reynolds divides by the actual nonzero group order. If one instead allows an empty prescribed set, the corresponding single condition is solved by the constant zero or one polynomial, and both empty sets allow zero. The construction uses finite products and least coordinate indices, with no AC.