Say that two structures $\mathfrak{A},\mathfrak{B}$ in finite languages $\Sigma,\Pi$ are parametrically equivalent - and write "$\mathfrak{A}\approx\mathfrak{B}$" - iff they have the same underlying set and the same set of definable-with-parameters relations.

I'm curious about the possible automorphism groups of a parametric equivalence class. Specifically, for a structure $\mathfrak{A}$ let $$\mathsf{AGS}(\mathfrak{A})=\{\operatorname{Aut}(\mathfrak{B}):\mathfrak{B}\approx\mathfrak{A}\}.$$ (It's here that the "same underlying set" condition is crucial: $\mathsf{AGS}(\mathfrak{A})$ is a family of subgroups of the full permutation group of that underlying set, and I am not identifying isomorphic elements of $\mathsf{AGS}(\mathfrak{A})$.)

For example, many rigid structures are parametrically equivalent to structures with $1$-transitively-acting automorphism group. A bit more surprisingly (to me at least), at MSE user Harry West showed that $\mathsf{AGS}(\mathfrak{A})$ need not be upwards-directed; however, the example he constructed was somewhat artificial.

But that's about all I know. In particular, I don't know much at all about computing $\mathsf{AGS}(\mathfrak{A})$ even for relatively simple $\mathfrak{A}$. I'd like to rectify this. To narrow the field (hehe), I'll focus on the following two candidates:

• The field of rationals $\mathfrak{Q}=(\mathbb{Q};+,\times)$.

• The field of reals $\mathfrak{R}=(\mathbb{R};+,\times)$.

Of course model-theoretically, $\mathcal{Q}$ is quite wild (e.g. bi-interpretable with the integers) while $\mathcal{R}$ is quite tame (e.g. decidable and o-minimal); it's not clear to me which of tameness or wildness makes $\mathsf{AGS}$ easier to analyze, but I suspect that at least one of these has some easily-establishable "coarse" properties (originally I just asked about $\mathfrak{Q}$).

Problem: describe at least one of $\mathsf{AGS}(\mathfrak{Q})$ or $\mathsf{AGS}(\mathfrak{R})$.

I'm especially interested in the question of whether each is upwards-directed.