Commuting permutations have separate sets of fixed points Lemma: If $\sigma, \tau \in S_{n}$ are commuting, then $\tau$ permutes elements $j \in \{1, 2, .., n\}$ that are left fixed by $\sigma$. 
Remark: There is a standard way of showing this in a group theory course and I am not asking to see this.
However, I was hoping to turn this into a linear algebra problem: 
Define $T_{\sigma}: \mathbb{C}^{n} \rightarrow \mathbb{C}^{n}$
by $T_{\sigma}(x_{1}, ..., x_{n}) = (x_{\sigma(1)},..., x_{\sigma(n)} )$, which is actually a linear transformation. 
Note $\sigma \circ \tau = \tau \circ \sigma \implies T_{\sigma \circ \tau } = T_{\tau \circ \sigma}$, $T_{\sigma} \circ T_{\tau} = T_{\tau} \circ T_{\sigma}$.
As commuting linear transformations, $T_{\tau}$ and $T_{\sigma}$ have a common eigenvector. This is the precise place where I want to deduce the original result, which I haven't been able to do so far. The entire attempt may be an exercise in futility. I will appreciate any ideas. 

You cannot deduce the fixed point property from the fact that $T_{\tau}$ and $T_{\sigma}$ have a common eigenvector, since all permutation matrices have the common eigenvector $(1,\ldots,1)$ but not all permutation matrices have pairwise disjoint sets of fixed points. Nice idea, though :-)
