Identity Permutation is Disjoint from All

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

Let $$e \in S_n$$ be the identity permutation on $$S_n$$.

Then $$e$$ is disjoint from every permutation $$\pi$$ on $$S_n$$ (including itself).

Proof
By definition of the identity permutation: $$\forall i \in \N^*: e \left({i}\right) = i$$.

Thus $$e$$ fixes all elements of $$S_n$$.

Thus each element moved by a permutation $$\pi$$ is fixed by $$e$$.

The set of elements moved by $$e$$ is $$\varnothing$$, so the converse is true vacuously.