Say that structures $\mathfrak{A},\mathfrak{B}$ with the same underlying set are **parametrically equivalent** iff every primitive relation/function in one is definable (with parameters) in the other. For example, every group $\mathfrak{G}=(G;\*,{}^{-1})$ is parametrically equivalent to its "torsor reduct" $\mathfrak{T}\_\mathfrak{G}=(G; (x,y,z)\mapsto x\*y^{-1}\*z)$.


Parametrically equivalent structures can still have very different combinatorial properties. In analogy with the notion of vertex transitive graphs, say that a structure $\mathfrak{A}$ is **point-transitive** iff the natural action of its automorphism group is $1$-transitive - that is, if for every $a,b\in\mathfrak{A}$ there is some $f\in Aut(\mathfrak{A})$ with $f(a)=b$. If $\mathfrak{G}$ is a nontrivial group then $\mathfrak{G}$ is not point-transitive *(no automorphism can move the identity)* but $\mathfrak{T\_G}$ is point-transitive *(for each $g\in G$ the map $a\mapsto g\*a$ is in $Aut(\mathfrak{T\_G}$))*.


I'm broadly interested in understanding the information captured by the family of automorphism groups of parametric equivalents of a given structure (see also [here](https://math.stackexchange.com/questions/4213906/must-the-poset-of-automorphism-group-variants-be-upwards-directed)). Motivated by the group/torsor example, one question which seems like it should be easy to answer is: which structures have a parametric equivalent whose automorphism group acts $1$-transitively? But even for simple structures things aren't very clear to me. So I'd like to start with a simple example:



> 
> Is the field of rationals $\mathfrak{Q}=(\mathbb{Q};+,\times)$ parametrically equivalent to a point-transitive structure?
> 
> 
> 


Per the torsor example above, the answer is affirmative for $(\mathbb{Q};+)$. It's also affirmative for $(\mathbb{Q};+,<)$ and various similar expansions, by the same construction, so the existence of such a parametric equivalent isn't connected to any obvious model-theoretic tameness property. However, multiplication seems to complicate things.



Is $\mathbb{Q}$ "equivalent" to a structure with transitive automorphism group action?
Consider $(\mathbb{Q},R)$, where $R$ is the 6-ary relation defined by $$R(a,b,c,d,e,f) \iff (a-b)(c-d)=(e-f)$$


The automorphism group of this structure includes at least the translations $x\to x+h$. Since those translations include $x \to x+(b-a)$, there is always a translation taking $a$ to $b$, so the group of automorphisms is point-transitive.


We can define $x+y$ and $xy$ in terms of $R$ as the unique $s$ and $t$ for which $R(1,0,s,x,y,0)$ and $R(x,0,y,0,t,0)$ respectively -- i.e. using the equations
\begin{align}
(1-0)(s-x)&=(y-0)\\
(x-0)(y-0)&=(t-0)
\end{align}


Finally, we can define $R$ in terms of $+$ and $\times$ directly by $ac+bd+f=ad+bc+e$.


