This is off a study guide for an exam I have in about two days. I really don't understand the problem entirely and would appreciate and help.
Let $\sigma \in S_{n}$, where $\sigma$ is a permutation and $S_{n}$ is the symmetric group. Let $x=\sum_{i=1}^{n} a_{i}e_{i} \in \mathbb{R}^{n}$ where $e_{i}$ are the standard basis elements of $\mathbb{R}^{n}$ and $a_{i}$ are scalars. Define a function $f_{\sigma}:\mathbb{R}^{n}\rightarrow \mathbb{R}^{n}$ by $f_{\sigma}(x)=\sum_{i=1}^{n}a_{i}e_{\sigma(i)}$
I'm asked to show that $f_{\sigma}$ is invertible and for $N>0$, $f_{\sigma}^{N}=I_{n}$
I know a permutation has a certain order, say $r$, so something like $\sigma^r$ is the identity permutation. I feel like that's something that has to be used, but I really do not know how to proceed. I'm totally stumped.
Note that $\sigma^{-1} \in S_n$ and if $x=\sum_{i=1}^na_ie_i\in \mathbb R^n$ then $y=\sum_{i=1}^na_ie_{\sigma^{-1}(i)} \in \mathbb R^n$ and $f_{\sigma}(y)=x$, hence $f$ is onto. Moreover if $f_{\sigma}(x_1)=f_{\sigma}(x_2)$ for$x_1=\sum_{i=1}^na_ie_i, x_2=\sum_{i=1}^nb_ie_i\in \mathbb R^n$ then $\sum_{i=1}^na_ie_{\sigma(i)}=\sum_{i=1}^nb_ie_{\sigma^(i)}$ which gives $a_i=b_i$ for each $i=1,\ldots,n$. Thus $x_1=x_2$ hence $f_{\sigma}$ is one to one. So, $f_{\sigma}$ is invertible.
On the other hand note that $f_{\sigma}^2(x)=\sum_{i=1}^na_ie_{\sigma^{2}(i)}$, and use $\sigma^{n!}=I$ for each $\sigma \in S_n$ to observe $f_{\sigma}^{n!}=I_n$.