Equivalence Class of Fixed Element

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 $i \in \operatorname{Fix} \left({\sigma}\right)$ if and only if $\left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)} = \left\{{i}\right\}$.

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

Suppose that $i \in \operatorname{Fix} \left({\sigma}\right)$.

Let $j \in \left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)}$.

Then by Condition for Membership of Equivalence Class and Permutation Induces Equivalence Relation:
 * $j \in \left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)} \iff i \mathrel{\mathcal R_\sigma} j \implies \exists m \in \Z: \sigma^m \left({i}\right) = j$

And by Fixed Point of Permutation is Fixed Point of Power:
 * $\sigma^m \left({i}\right) = i \implies i = j$

Therefore $\left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)} \subseteq \left\{{i}\right\}$ and so $\left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)} = \left\{{i}\right\}$.

For the converse, suppose that $\left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)} = \left\{{i}\right\}$.

It is easily seen that $i \mathrel{\mathcal R_\sigma} \sigma \left({i}\right) \iff \sigma \left({i}\right) \in \left[\!\left[{i}\right]\!\right]_{\mathcal R \left({\sigma}\right)}$.

Therefore $\sigma \left({i}\right) = i \implies i \in \operatorname{Fix} \left({\sigma}\right)$

Hence the result.