I am now confused with such problem as title goes. To be exact, the problem is
Does there exist a functor from $A:\mathsf{Field}\to \mathsf{Field}$ with a natural transformation from identity functor $\iota: \operatorname{id}\to A$ such that for each $F$, $A(F)$ is the algebraically closure of $F$ through $\iota_F:F\to A(F)$?
It is not easy rather than first glimpse. Let me explain.
Note that, the existence of algebraic closure only ensures that there exist a map from $\operatorname{Obj}(\mathsf{Field})$ to itself. Since the "extension" property is not unique, it is not generally true that we can extend the map to $\operatorname{Mor}(\mathsf{Field})$ for arbitrary choice of algebraic closure.
-
For example, consider the fields
$$\begin{array}{ccc}
\mathbb{Q}[\sqrt[3]{2}, \sqrt{2}] &\to & \mathbb{Q}[\omega\sqrt[3]{2}, \sqrt{2}]\\
\uparrow &&\uparrow \\
\mathbb{Q}[\sqrt[3]{2}] & \to & \mathbb{Q}[\omega\sqrt[3]{2}]
\end{array}$$
If we choose the algebraically closure of $\left[\begin{matrix}\mathbb{Q}[\sqrt[3]{2}, \sqrt{2}] & \mathbb{Q}[\omega\sqrt[3]{2}, \sqrt{2}]\\ & \mathbb{Q}[\omega\sqrt[3]{2}]\end{matrix}\right]$ by inclusion to $\overline{\mathbb{Q}}$, and the closure of $\mathbb{Q}[\sqrt[3]{2}]\to \overline{\mathbb{Q}}$ by $\sqrt[3]{2}\mapsto \omega\sqrt[3]{2}$.
We cannot extend a well-defined functor. Similar problem exists for transcendental extension, for example, square like this
$$\begin{array}{ccc}
\mathbb{C}[X,Y] &\to & \mathbb{C}[X^2,Y]\\
\uparrow &&\uparrow \\
\mathbb{C}[X] & \to & \mathbb{C}[X^2]
\end{array}$$
A reasonable method is to avoid phenomenon above is as follow.
Fix an algebraically closed field $F$, and take all of its subfields as "skeleton", then fix an isomorphism to a subfields of $F$ from all fields whose algebraic closure is $F$ up to an isomorphism. The isomorphic class of algebraically closure are completely dependen by its characteristic and the transcendental dimension over prime field $\mathbb{Q}$ or $\mathbb{F}_p$. -
Now the problem is how to naturally chose extensions for endmorphisms. But unfortunately, the choice is fragile. For instance, consider the following diagram
$$\begin{array}{ccccl}
\mathbb{Q}[\sqrt{3}, \sqrt{2}] &\to & \mathbb{Q}[\sqrt{3}, \sqrt{2}] &: &\sqrt{3}\mapsto -\sqrt{3},\sqrt{2}\mapsto \pm \sqrt{2}\\
\uparrow &&\uparrow \\
\mathbb{Q}[\sqrt{3}] & \to & \mathbb{Q}[\sqrt{3}] &:&\sqrt{3}\mapsto -\sqrt{3}
\end{array}$$
There is no suitable choice such that
$$\begin{array}{ccccl}
\overline{\mathbb{Q}} &\to & \overline{\mathbb{Q}} &: &\sqrt{3}\mapsto -\sqrt{3},\sqrt{2}\mapsto \pm \sqrt{2}\\
\parallel &&\parallel \\
\overline{\mathbb{Q}}& \to & \overline{\mathbb{Q}} &:&\sqrt{3}\mapsto -\sqrt{3}
\end{array}$$
commutes for both $\pm=+$ and $\pm=-$.
Best Answer
No, this is not possible. For instance, let $K$ be any field with a automorphism $f:K\to K$ whose order is finite and greater than $2$. Then $A(f):A(K)\to A(K)$ would be an automorphism of the same order extending $f$. But no such automorphism exists: by the Artin-Schreier theorem, any finite-order automorphism of an algebraically closed field has order at most $2$.
Or without using any big theorems, you can find problems just looking at finite extensions. For instance, if $f$ is the Frobenius automorphism of $\mathbb{F}_{p^2}$ then $F(f)$ is an extension to an algebraic closure which still has order $2$. Since $\mathbb{F}_{p^4}$ is normal over $\mathbb{F}_{p}$, $F(f)$ restricts to an automorphism of $\mathbb{F}_{p^4}$, which must be the Frobenius squared in order to have order $2$. But the Frobenius squared does not restrict to $f$ on $\mathbb{F}_{p^2}$, so this is a contradiction.