Equivalence Class of Fixed Element/Corollary

Theorem
Let $S_n$ denote the symmetric group on $n$ letters.

Let $\sigma \in S_n$.

Let $\mathcal R_\sigma$ be the equivalence defined in Permutation Induces Equivalence Relation.

Let $i \in \N^*_{\le n}$.

Then $i \notin \operatorname{Fix} \left({\sigma}\right)$ if and only if $\left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)}$ contains more than one element.

Proof
From Equivalence Class of Fixed Element and Biconditional Equivalent to Biconditional of Negations, $i \notin \operatorname{Fix} \left({\sigma}\right) \iff \left\{{i}\right\} \ne \left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)}$.

Because the Biconditional is Transitive, it suffices to show that $\left\{{i}\right\} \ne \left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)}$ if and only if $\left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)}$ contains more than one element.

Suppose that $\left\{{i}\right\} \ne \left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)}$.

From the definition of an equivalence relation it is easily seen that:
 * $\left\{{i}\right\} \subseteq \left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)}$

And from the hypothesis:
 * $\left\{{i}\right\} \subset \left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)}$

Therefore $\left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)}$ contains more than one element.

Conversely, if $\left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)}$ contains more than one element, then obviously $\left\{{i}\right\} \ne \left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)}$.