Assume a finite field extension $\mathbb{C}/K$ such that $[\mathbb{C}:K]>2$. Let $\varphi \in \text{Aut}(\mathbb{C}/K)$, so $\varphi \in \text{Aut}(\mathbb{C})$ and $\varphi\vert_K=\text{id}\vert_K$.
Is it possible to show one of the three:
i. that the image $\varphi(\mathbb{R})$ is a subset of $\mathbb{R}$.
ii. that $\varphi$ is also a $\mathbb{R}$-Automorphism (which would follow from i.).
iii. that $\varphi$ is continuous on $\mathbb{R}$ (which follows from either i. or ii.).
We know that $(\varphi(i))^2=\varphi(i^2)=\varphi(-1)=-1$ as $\varphi\vert_{\mathbb{Q}}=\text{id}_{\mathbb{Q}}$ and we therefore have either $\varphi(i)=i$ or $\varphi(i)=-i$.
We also know that $\varphi$ fixes $K\cap\mathbb{R}$ pointwise and due to Artin-Schreier we have that $i$ is not in either $K$ or $K\cap\mathbb{R}$ as $K \cap \mathbb{R}$ is a subfield of $K$ and $\mathbb{R}$.
As $\mathbb{C}$ has characteristic $0$, so does $K$ and $K\cap\mathbb{R}$, and therfore $\mathbb{Q} \subseteq K\cap\mathbb{R} \subseteq K \subset \mathbb{C}$.
I tried to experiment with $K\cap\mathbb{R}$ but quite unsuccesful, as it could be equal to $\mathbb{Q}$ which isn#t of much use. It's also to note, that $\mathbb{C}/K$ is a Galois-Extension.
In the first version of the post I forgot to mention, that the degree of $\mathbb{C}/K$ is supposed to be finite!
Thanks for checking in :)
As linked to in a comment, $\mathbb{C}$ has "wild" automorphisms over $\mathbb{Q}$ or really any subfield of $\mathbb{C}$ of transcendence degree at most countable. These are basically impossible to describe, but they definitely don't satisfy any of these properties you mentioned.
Any field automorphism must preserve $\mathbb{Q}$ pointwise, so if it were continuous on $\mathbb{R}$ it would have to be the identity on $\mathbb{R}$ and therefore the automorphism of $\mathbb{C}$ would have to be either the identity map or complex conjugation.