Let $F$ be a field and $f(x_1,\ldots,x_n), g(x_1,\ldots,x_n) \in F[x_1,\ldots, x_n]$. Suppose that $h(x_1,\ldots,x_n):=f(x_1,\ldots,x_n)/g(x_1,\ldots, x_n)$ is symmetric, in the sense that for every $\sigma \in S_n$, $\Phi(\sigma)(h) = h$, where $\Phi(\sigma)$ is the unique automorphism of $F(x_1,\ldots,x_n)$ which extends the automorphism $\phi(\sigma)$ of $F[x_1,\ldots,x_n]$ which fixes $F$ and sends $x_i$ to $x_{\sigma(i)}$. Suppose further that $f$ and $g$ have no common factor.
How to prove that both $f$ and $g$ are symmetric? I tried hard on this one but no ideas come to mind.
HINT:If $f/g$ is in lowest terms and $\frac{\sigma f}{\sigma g} = \frac{f}{g}$ for all $\sigma$ then $\sigma g$ divisible by $g$ for all $\sigma$. Taking $\sigma \mapsto \sigma^{-1}$ we get that $\sigma g = \epsilon_{\sigma} g$ for all $\sigma$, where $\sigma \mapsto \epsilon_{\sigma}$ is a ($1$-) character of $S_n$. If $\epsilon_{\sigma} \equiv 1$ then $g$ symmetric, done. Otherwise, $\epsilon_{\sigma}$ is the signature. But then $g$ is skew-symmetric, and then, necessarily, $f$ is also skew symmetric ( since their quotient is symmetric). But every skew-symmetric polynomial is divisible by $\delta=\prod_{i< j}(x_i - x_j)$, and thus $f$, $g$ are not relatively prime, contradiction.
Obs: It may be convenient to produce explicit symmetric polynomial as a quotient $\frac{f}{\delta}$, with $f$ skew-symmetric, $\delta$ as above, ( see Schur functions), although the fraction is clearly not in lowest terms.