If $\sigma(\alpha)$ takes finitely many values when $ \sigma$ runs over automorphism (as field) of $ \Bbb{C}$, why $ \alpha$ is algebraic number?

Question

Let $ \alpha$ be complex number. If $\sigma(\alpha)$ takes finitely many values when $ \sigma$ runs over automorphism (as field) of $ \Bbb{C}$, why $ \alpha$ is algebraic number ?

To prove this, firstly I want to prove $\Bbb{Q}( \alpha)/\Bbb{Q}$ is finite extension. Thank you in advance.