Extending an automorphism of finite order of $A \subset \mathbb{C}$ to an automorphism of the same order

Let $A \subset \mathbb{C}$ be a subfield of $\mathbb{C}$.

According to a result of P.B. Yale, Theorem 7,
any automorphism of $A$ can be extended to an automorphism of $\mathbb{C}$;
see also this question (and its good comments and answers).

My question: If we know, in addition, that a given automorphism of $A$ is of finite order (for example, of order $2$, namely an involution),
is it possible to know that its extension to $\mathbb{C}$ is of the same order
(or at least also of finite order)?
Or is it hopeless, and the best we can obtain is ‘just’ that it has an extension to $\mathbb{C}$, probably of infinite order?

In other words, can we adjust Zorn’s Lemma to guarantee a finite order extension? (I am afraid that the answer is no.. I hope I am wrong).

Thanks for any comments.

Solutions Collecting From Web of "Extending an automorphism of finite order of $A \subset \mathbb{C}$ to an automorphism of the same order"

It is usually not possible to extend a finite order automorphism of $A$ to an automorphism of $\mathbb{C}$ of finite order. More precisely, it follows from the Artin-Schreier theorem that any nontrivial finite order automorphism of $\mathbb{C}$ has order $2$, maps $i$ to $-i$, and has a fixed field which is real closed. So if an automorphism of $A$ has order greater than $2$, it cannot have any finite order extension to $\mathbb{C}$. Even if an automorphism of $A$ has order $2$, it may not extend to a finite order automorphism of $\mathbb{C}$ (for instance, if $i\in A$ and the automorphism fixes $i$).