Suppose  E \leq \bar{F} is an algebraic extension of of  F and  \alpha, \beta \in E are conjugate over  F . Then the conjugation isomorphism  \psi_{\alpha, \beta}: F(\alpha) \to F(\beta) can be extended to an isomorphism of  E onto a subfield of  \bar{F} .

So is it never always possible to extend it onto  \bar{F} itself? Is  \mathbb{C} its own algebraic closure? In other words, does  \mathbb{C} = \bar{\mathbb{C}} ?