A classic result says the automorphism group of $\mathbb{R}$ (over $\mathbb{Q}$) is trivial. The proof is simple: every automorphism preserves squares, and hence fixes the positive reals, so it must be order preserving. Since it must fix $\mathbb{Q}$, and $\mathbb{Q}$ is dense in $\mathbb{R}$, if any real number were not fixed, this would yield a contradiction.

In larger real-closed fields where $\mathbb{Q}$ is not dense, automorphisms are still order preserving, but the argument that they are trivial does not work. I haven't found any examples of a non-trivial automorphism of a real-closed field, but I also can't prove they don't exist. So, the first question I'd like to ask is whether all automorphism groups of real-closed fields are trivial.

If they aren't all trivial, then I want to know what, if anything, we can say. To ask a less vague question, can we classify the automorphisms of the hyperreals? How many are there, and what is their groups structure like?

Models with second order properties IV. A general method and eliminating diamonds, 1983 sciencedirect.com/science/article/pii/0168007283900131) $\endgroup$ – YCor Dec 31 '16 at 7:47order-preservingone of any subfield, but those are all trivial. Only nonarchimedean real-closed fields may admit nontrivial automorphisms. $\endgroup$ – Emil Jeřábek Dec 31 '16 at 14:50uniqueordering whose positive elements are precisely the sums of squares, and since an automorphism will then preserve the ordering, its restriction to the set of roots $r_1 < r_2 < \ldots < r_n$ of an $f$ must take $r_i$ to $r_i$. See Lang's Algebra, chapter XI section 2. $\endgroup$ – Todd Trimble♦ Dec 31 '16 at 14:50