If $\frac f g$ is symmetric then what conclusion can we make about $f$ and $g$?

Question

Let $K$ be a field (or a commutative ring with identity). Consider $f,g \in K[X_1,X_2, \cdots , X_n]$ with $g \neq 0.$ Suppose that $\frac f g \in K \left (X_1,X_2, \cdots , X_n \right)$ is symmetric i.e. $\frac f g \in \text {Fix}_{S_n} K\left (X_1,X_2, \cdots, X_n \right )$ where $S_n$ denotes the symmetric group on $n$-symbols and $$\text {Fix}_{S_n} K\left (X_1,X_2, \cdots, X_n \right ) = \left \{h \in K \left (X_1,X_2, \cdots , X_n \right )\ \big |\ \sigma (h) = h\ \text {for all}\ \sigma \in S_n \right \}.$$

What conclusion can we make about $f$ and $g$ from here? Are they necessarily symmetric? If so why?

Any help regarding this will be highly appreciated. Thank you very much.

Answer 1

I have found the answer. It's quite easy. Take for instance $f=X_1^2X_2 \in K[X_1,X_2]$ and $g=X_1 \in K[X_1,X_2].$ Neither $f$ nor $g$ is symmetric. But $\frac f g = X_1X_2 \in K [X_1,X_2] \subseteq K \left (X_1,X_2 \right )$ is indeed symmetric.