Let $K$ be a field with an involution $*$, meaning $*:K\to K$ is an automorphism and $(x^*)^*=x$ for all $x\in K$. Suppose further that the fixed field of $*$ is ordered (i.e., it can be given an ordering that is compatible with the field structure).
Examples are $\mathbb R$ (with the identity map) and $\mathbb C$ (with complex conjugation). Are there any interesting nontrivial examples other than these? By interesting, I mean not just subfields of $\mathbb C$, whose ordered fixed fields are subfields of $\mathbb R$.
Background:
I'm asking because these are the properties that enable the definition of inner products on real and complex vector spaces, and I'm curious about generalizing this definition. So in this context it is natural to consider fields with the properties given above. But these seem like quite constraining properties, so I'd like to see whether there is really anything interesting to be gained from this generalization.
Similar previous question:
I asked a similar question here, except that the "ordered subfield" condition was different, specifically asking that that the subfield be identified with $\mathbb R$. But now I think this is the more interesting (and precise) question, and it was mentioned but not answered in the comments of the linked post.
Try the Puiseux series $E=\bigcup_{n\ge 1}\Bbb{C}((x^{1/n})), F=\bigcup_{n\ge 1}\Bbb{R}((x^{1/n}))$ so that $E=F+iF, a+ib\to a-ib$ is the involution. By the axiom of choice (algebraically closed field of same cardinality) $E$ is isomorphic to $\Bbb{C}$. $F$ is ordered ($\sum_{k\ge -K} a_k x^{k/n}>0$ iff the first non-zero $a_k$ is $>0$). It is not isomorphic to $\Bbb{R}$ nor to a subfield of $\Bbb{R}$ (as $1+mx$ has a square root in $F$ for all $m\in \Bbb{Z}$).