Composite of Automorphisms is Automorphism

Theorem
Let $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$ be an algebraic structure.

Let:
 * $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$
 * $\psi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$

be automorphisms.

Then the composite of $\phi$ and $\psi$ is also an automorphism.

Proof
From Composite of Homomorphisms on Algebraic Structure is Homomorphism, $\psi \circ \phi$ is a homomorphism.

By the definition of a composite mapping, $\psi \circ \phi$ is a mapping from $S$ into $S$.

Hence $\psi \circ \phi$ is an automorphism.