Algebraically closed fields with only definable automorphisms According to the paper "Models with second order properties IV. A general method and eliminating diamonds" by Shelah, he constructs ordered fields with only definable automorphisms.
I haven't read the paper yet, but am interested in the possibility of constructing algebraically closed fields with only definable automorphisms, if my understanding of the abstract of the above paper is correct.
So my question is: Is there a possibility of constructing algebraically closed fields with only definable automorphisms?
 A: Such fields do not exist by simple cardinality considerations: an algebraically closed field of size $\kappa$ has at most $\kappa$ definable automorphisms, whereas it is easy to see that it has $2^\kappa$ automorphisms in total (e.g., if $\kappa=\aleph_0$, the automorphism group is a perfect Polish space, and if $\kappa>\aleph_0$, the field has a transcendence basis of size $\kappa$ every permutation of which extends to an automorphism).
In fact, definable automorphisms of algebraically closed fields are very limited, and one can list them all.

Proposition 1: The only definable automorphism of an algebraically closed field $K$ of characteristic $0$ is the identity.

Proof: If $\sigma$ is a definable automorphism, the fixed field $F$ of $\sigma$ is a definable subset of $K$, hence it is finite or cofinite as $K$ is a minimal structure. However, every proper subfield of $K$ is an infinite set with an infinite complement, hence the only possibility is $F=K$, i.e., $\sigma=\mathrm{id}$.

Proposition 2: If $K$ is an algebraically closed field of characteristic $p>0$, the only definable automorphisms of $K$ are the iterates of the Frobenius automorphism $\tau_n(x)=x^{p^n}$, where $n\in\mathbb Z$.

Proof: Let $\sigma$ be an automorphism definable from parameters $\vec a\in K$. Passing to an elementary extension of $K$ if necessary, we may assume there is an element $u\in K$ transcendental over $F=\mathbb F_p(\vec a)$. Every automorphism fixing $F(u)$ must also fix $\sigma(u)$, hence $\sigma(u)$ is algebraic and purely inseparable over $F(u)$, i.e.,
$$\sigma(u)^{p^k}=\frac{f(u)}{g(u)}$$
for some $k$ and nonzero polynomials $f,g\in F[x]$. It is enough to show that the definable automorphism $\tau_k\circ\sigma$ equals some $\tau_n$, hence we may assume $k=0$ WLOG. Since $u$ can be mapped to any element transcendental over $F$ by an automorphism fixing $F$, the identity
$$\sigma(x)g(x)=f(x)$$
must hold for all such transcendental elements $x$; using the minimality of $K$, it actually holds for all but finitely many elements of $K$, and by multiplying $f$ and $g$ by linear factors, we can ensure that it holds for all elements of $K$.
For any $m>0$, $\sigma$ restricts to an automorphism of $\mathbb F_{p^m}$, hence it coincides on $\mathbb F_{p^m}$ with $\tau_n$ for some $0\le n<m$. This means that the polynomial
$$f(x)-x^{p^n}g(x)$$
vanishes on $\mathbb F_{p^m}$. If $m$ is sufficiently large, its degree is less than $p^m$, hence it is in fact the zero polynomial. This means that
$$\sigma(x)=x^{p^n}=\tau_n(x)$$
holds for all but finitely many $x\in K$, which implies $\sigma=\tau_n$.
