I have the following problem that I believed I solved but am trying to understand better.
Let $n\geq 2$. Consider $\sigma \in S_{n}$ and $f \in \mathbb{R}[T_1,\dots , T_n]$, let $\sigma(f)=f(T_{\sigma(1)},\dots,T_{\sigma(n)})$. Show that for any $f$, the number of different polynomials of the form $\sigma(f)$ divides $n!$
It is easy to show this is a group action of $S_n$ acts on $\mathbb{R}[T_1,\dots,T_n]$. $\\ $ Since, $i(f)=f(T_{i(1)},...,T_{i(n)})=f(T_1,\dots,T_n)$ and $\sigma(\tau(f))=(\sigma\tau)(f)$
This is what I got, let $g \in \mathbb{R}[T_1,\dots,T_n]$, then the number of polynomials of the form $\sigma(f)$ is the orbit of $g$? Then, $\left\vert{S_{n}g}\right\vert = [Sn:S_{ng}]$.But $S_{ng}=\{\sigma \in S_{n}|\sigma(f)=f(T_1,\dots T_n)\}$. Then, $\left\vert{S_{ng}}\right\vert = 1$. So then $\left\vert{S_{n}g}\right\vert=n!$
My question is, am I right in interpreting "the number of different polynomials of the form $\sigma(f)$" as being the set $Gx=S_{n}g$, or in other words the orbit of $g$ in $S_n$?
It looks like you're getting bogged down in notation but have roughly the right idea. If I've interpreted the problem correctly, you're asked to fix a polynomial $f(T_1,\ldots,T_n) \in \mathbb{R}[T_1,\ldots,T_n]$ and show that $$ \#\{\sigma(f) : \sigma \in S_n\} \mid n! $$ (so $f$ is fixed and $\sigma$ varies). Then the shortest hint is to apply the Orbit-Stabilizer Theorem. If you haven't seen that theorem yet, the following line of inquiry shows you how to basically prove it from scratch in this case (and for finite groups in general mutatis mutandis):
You can then use 2. to define an equivalence relation $\sim$ on $S_n$ by saying that $$\sigma \sim \tau \iff \text{[whatever you found was true when $\sigma(f) = \tau(f)$]}$$ and using this and general properties of equivalence relations to conclude that $\#\{\sigma(f) : \sigma \in S_n\} = [G:H]$.