Show $v\in FS_n$ is an $F$-multiple. This is coming from Exercise 8 in Section 18.1 of Dummit and Foote.  We are talking about representation theory and in particular focusing on Example 3 and 10 in this section.  
Let $n\in \mathbb{Z}^+$ and $G=S_n$ and $V$ be an n-dimensional vector space over $F$ with basis $e_1, e_2,...,e_n$.  Let $S_n$ act on $V$ by defining for all $\sigma\in S_n$ that $\sigma e_i=e_{\sigma (i)} $ for all $1\leq i\leq n$. So $\sigma$ acts by permuting the subscripts of the basis elements.  In the book they show that $V$ is an $FS_n$-module, where $FS_n$ is a group ring over $F$.    
Our exercise says: Let $V$ be an $FS_n$-module.  Suppose $v\in V$ such that $\sigma  v = v$ for all $\sigma\in S_n$. We want to show that $v$ is an $F$-multiple of $e_1 +...+e_n$.
Now, I'm not completely sure what an $F$-multiple is, but I'm assuming that it means we want to show $v=\alpha (e_1+...+e_n)$ for $\alpha \in F$.  
I've also just started working with representation theory so I don't really know what I'm doing, but here is what I have so far:
Suppose $v\in FS_n$ and $\sigma v=v$ for all $\sigma\in S_n$.  Let $v=\alpha _1e_1+...+\alpha _n e_n$.  We want to show that $\alpha _1=...=\alpha _n$.  Then we would have that $v$ is an $F$-multiple.  So far I have that since $\sigma v=v$ then $\sigma (\alpha _1e_1+...+\alpha _n e_n)=\alpha _1e_1+...+\alpha _n e_n$.  By definition of our action, we have that $\alpha _1e_{\sigma (1)}+...+\alpha _n e_{\sigma (n)}=\alpha _1e_1+...+\alpha _n e_n$.  Therefore, $\alpha _ie_{\sigma (i)} = \alpha _ie_i$ for all $i$. Hence, $e_{\sigma (i)} = e_i$ for all $i$.   So, $\sigma (i) = i$ for all $i$.  So, $\sigma $ is the identity permutation. 
This gives us that $\sigma v=v$ implies $\sigma$ is the identity permutation.  
This is all I have so far.  I don't know how this implies $\alpha _1=...=\alpha _n$.  This might not even be how we prove this, but this is all I could come up with.  
 A: Suppose that $\sigma v=v$ for all $\sigma$, and write $v=\alpha_1e_1+\dots+\alpha_ne_n$. As you said, it is enough to show that $\alpha_1=\dots=\alpha_n$.
If $\sigma=(1 2)$, then $\sigma v=\alpha_2e_1+\alpha_1e_2+\alpha_3e_3+\dots+\alpha_ne_n$, hence
$$ 0=\sigma v-v=(\alpha_2-\alpha_1)e_1+(\alpha_1-\alpha_2)e_2$$
and since $e_1$ and $e_2$ are linearly independent, this implies that $\alpha_1=\alpha_2$. The same argument with $\sigma=(1\,j)$ shows that $\alpha_j=\alpha_1$ for all $j$.
A: Your interpretation is correct and strategy is good but perhaps you are overthinking the execution. 
Let i and j be two arbitrary basis indices. Pick the symmetry $\sigma$ that interchanges $e_i$ and $e_j$ and leaves the other e's fixed. The equation $\sigma v=v$ just means that you can interchange the coefficients of $e_i$ and $e_j$ and you have two representations of $v$. But by uniqueness of representation in terms of the basis, this means $\alpha_i=\alpha_j$. So, the coefficients are all pairwise equal, hence mutually equal.
