Group Action of Symmetric Group Acts Transitively

Theorem
Let $S$ be a set.

Let $\left({\Gamma \left({S}\right), \circ}\right)$ be the group of permutations on $S$.

Let $*: \Gamma \left({S}\right) \times S \to S$ be the group action defined as:
 * $\forall \pi \in \Gamma \left({S}\right), \forall s \in S: \pi * s = \pi \left({s}\right)$

Then $*$ is a transitive group action.

In other words, $\left({\Gamma \left({S}\right), \circ}\right)$ acts transitively on $S$ by $*$.

Proof
By Group Action of Symmetric Group, $*: \Gamma \left({S}\right) \times S \to S$ is indeed a group action

Let $s, t \in S$.

As $\Gamma \left({S}\right)$ is the group of permutations on $S$, there exists a permutation $\pi \in \Gamma \left({S}\right)$ such that:
 * $\pi \left({s}\right) = t$

This holds for any $s, t \in S$.

Thus:
 * $\forall t \in S: t \in \operatorname {Orb} \left({s}\right)$

and so $S$ consists of a single orbit.

Hence the result by definition of transitive group action.