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It is well-known that an arbitrary projective plane can have very different symmetry group to a field plane. In particular, the symmetries are not transitive on the set of fundamental quadrangles. This corresponds to algebraic properties of coordinatising planar ternary rings.

However, I'm wondering what the situation is for specified points when we have fewer of them. For instance, suppose we have three chosen points, not four. In the coordinatisation process this corresponds to picking an "origin" and "axes", but not a line of "slope 1". Two chosen points correspond to picking one of the "axes" and an "origin", and one chosen point corresponds to the "origin". In any of these cases, is it possible to map one choice onto any other possible choice by an automorphism of the plane? Please note I only care about the finite case here.

Suppose this is possible for finite planes. If we picked a pair of fundamental quadrangles, then a(n ordered) subset of $n$ points ($n=1,2,3$) in it, then to what do automorphisms mapping the one chosen subset to the other correspond? For $n=1$, presumably we get a function from one coordinatising planar ternary ring to the other that only preserved the zero element.

Edit As pointed out by Gordon Royle in the comments, this fails already for $n=1$. So the more nuanced question is: what do we know about the partition of the sets of points/pairs of points/triple of non-collinear points into orbits under automorphisms? And what does this tell us about the induced morphisms between coordinatising planar ternary rings?

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  • $\begingroup$ There are many planes that do not even have automorphism group transitive on points. $\endgroup$ Aug 5 at 22:53

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