Question:
Let $E/F$ be a normal extension,$\bar F$ be the algebraic closure of $F$.Denote the set of embeddings from $E$ to $\bar F$ fixing $F$ as $Hom_F(E,\bar F)$.
The question is can we construct a bijection between $Aut_F(E)$ and $Hom_F(E,\bar F)$?
Attempt:
Since $E/F$ is normal,if we assume that $E\subset \bar F$,then every embeddings $\Phi:E\to\bar F$ fixing $F$ satisfies $\Phi(E)=E$.
So restrict the image to $E$ we get $\phi\in Aut_F(E)$.
Conversely,since $E\subset \bar F$,there is a natural embedding from $E$ to $\bar F$(which is the inclusion $\iota$).So we can map $\phi\in Aut_F(E)$ to $\Phi=\iota\circ\phi\in Hom_F(E,\bar F)$,and the two maps are the inverse of each other.
But does a normal extension $E/F$ necessarily require $E\subset \bar F$?If we remove the requirement,I find it hard to construct such a map,since we may not have $\Phi(E)=E$ and the natural embedding.
Just fix one particular embedding $\Phi_0:E\to\overline{F}$ and let $E_0=\Phi_0(E)$. Then $E$ is isomorphic to $E_0$ via $\Phi_0$ and this gives bijections $\operatorname{Aut}_F(E)\to \operatorname{Aut}_F(E_0)$ and $\operatorname{Hom}_F(E,\overline{F})\to\operatorname{Hom}_F(E_0,\overline{F})$, which you can then just compose with the canonical bijection $\operatorname{Aut}_F(E_0)\to \operatorname{Hom}_F(E_0,\overline{F})$ that you have identified.
(Note that this means that in general, the bijection $\operatorname{Aut}_F(E)\to \operatorname{Hom}_F(E,\overline{F})$ is not canonical: it depends on a choice of embedding of $E$ into $\overline{F}$. The more canonical statement you can make is that the group $\operatorname{Aut}_F(E)$ acts on $\operatorname{Hom}_F(E,\overline{F})$ by composition, and this action is simple and transitive.)