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

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