= Solution
Use the realization $R_3=W\subset\mathbb C^4$ from <2/c/solution>. Put $x=(\phi_1,\phi_2,\phi_3,-\phi_1-\phi_2-\phi_3)$, which identifies the three-dimensional <scalar field> with $W$. The <polynomial invariants of the standard S4 representation> include the symmetric power sums
$$
I_2=\sum_{i=1}^4x_i^2,
\qquad I_3=\sum_{i=1}^4x_i^3,
$$
because every <permutation> merely rearranges their summands. In these coordinates,
$$
\boxed{I_2=\phi_1^2+\phi_2^2+\phi_3^2+(\phi_1+\phi_2+\phi_3)^2,\qquad
I_3=-3(\phi_1+\phi_2)(\phi_2+\phi_3)(\phi_3+\phi_1).}
$$
Neither <polynomial> vanishes identically: at $(\phi_1,\phi_2,\phi_3)=(1,1,1)$ they equal $12$ and $-24$. Hence coefficients multiplying $I_2$ and $I_3$ provide nonzero invariant quadratic and cubic terms. A change of <basis> transports these <polynomial invariants> to any equivalent realization of $R_3$. For a real <scalar field>, both are real. For a <complex scalar field>, a real <scalar potential> may contain $\lambda_k I_k+\lambda_k^*I_k^*$; the quadratic <Hermitian form> $\sum_i|x_i|^2$ is another invariant.
As an independent <character> check, the <symmetric square of a group representation> has <character> $[\chi(g)^2+\chi(g^2)]/2$, and the <symmetric cube of a group representation> has <character> $[\chi(g)^3+3\chi(g)\chi(g^2)+2\chi(g^3)]/6$. With $\chi=\chi_3$ these are $(6,2,2,0,0)$ and $(10,2,-2,1,0)$. Their <character inner products> with $\chi_0$ are both one, so each degree has exactly one independent <polynomial invariant>.
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