Let $k$ be an infinite field, perhaps take $k = \mathbb{C}$ if it simplifies matters.
I will be asking a question about $\mathbb{P}^2$ for definiteness and to simplify definitions/notations, but feel free to generalize to $\mathbb{P}^d$ if you can.
For $n\geq 4$, define a “ruler-only construction on $n$ points” to be a means of constructing, given $n$ points $P_0,\ldots,P_{n-1}$ in sufficiently general position in the plane $\mathbb{P}^2$ (over $k$), a point $Q$, by repeated (finite) iteration of the operations of drawing the line connecting two points or taking the intersection of two lines.
(In more fancy terms, a ruler-only construction is a rational map $\mathbb{P}^2 \times \cdots \times \mathbb{P}^2 \dashrightarrow \mathbb{P}^2$ with $n$ factors at the source, that is covariant under $\mathit{PGL}_3$ acting on each of the $\mathbb{P}^2$.)
Equivalently, it is well known that, fixing coordinates such that $P_0 = (1{:}0{:}0)$, $P_1 = (0{:}1{:}0)$, $P_2 = (0{:}0{:}1)$ and $P_3 = (1{:}1{:}1)$, we can see view a construction as being a point $Q$ in $\mathbb{P}^2$ with coordinates in the rational function field $k(x_4,y_4,\ldots,x_{n-1},y_{n-1})$ where $(1 : x_i : y_i)$ are the coordinates of $P_i$ (for $4\leq i\leq n-1$). However, this may not be the best way of putting things because this breaks the symmetry of the $P_i$ which is precisely what I am interested in.
Indeed, if $\sigma \in \mathfrak{S}_n$ is a permutation of $\{0,\ldots,n-1\}$, we can apply $\sigma$ to a construction $Q$, meaning that we define $Q^{\sigma}$ by applying the construction of $Q$ to the points $P_{\sigma(0)},\ldots,P_{\sigma(n-1)}$ (possibly restricting further so that they too are “sufficiently general” for the construction to work). Say that the construction¹ is invariant under $\sigma$ when $Q^{\sigma} = Q$ (for sufficiently general starting points). And of course, the symmetry group of the construction is the subgroup $H \leq \mathfrak{S}_n$ consisting of those $\sigma$ such that the construction is invariant under $\sigma$. (In particular, we then get $(\mathfrak{S}_n : H)$ distinct points $Q^{\sigma}$ by applying the construction to permutations of $P_0,\ldots,P_n$.)
- I guess I should more properly say that the “constructed point” is invariant because the actual steps of the construction need not be invariant. But since I am only interested in the constructed point, not the steps of the construction, this will do.
QUESTION: Given $n\geq 4$, can we completely describe the set of possible symmetry groups (as subgroups of $\mathfrak{S}_n$) of a ruler-only construction on $n$ points in the plane?
Remarks:
Even for $n=4$ I don't know the full answer. However, to illustrate the question, I can show that $\mathfrak{S}_4$ itself is not a possible symmetry group. Indeed, a ruler-only construction on $4$ points is simply given by a point in $\mathbb{P}^2(k)$ if we make $P_0,P_1,P_2,P_3$ into a standard basis as above. Now permutations of $\{0,1,2\}$ act as permutations of the three coordinates, so the only point invariant under those is $(\lambda:\lambda:\lambda) = P_3$, and since it is trivially not invariant under $\mathfrak{S}_4$, there are no fully $\mathfrak{S}_4$-invariant points.
For $n=8$, however, there is a fully $\mathfrak{S}_8$-invariant construction: given $P_0,\ldots,P_7$ eight points in $\mathbb{P}^2$ we can define $Q$ as the ninth common intersection point of all cubics passing through $P_0,\ldots,P_7$: this can be written as a ruler-only construction (see here and the references therein) and it is, of course, completely symmetric in $P_0,\ldots,P_7$.
I don't even know whether we can always make a construction “less symmetric”, i.e., if $H' \leq H \leq \mathfrak{S}_n$ and $H$ is a possible symmetry group of a construction on $n$ points, is it true that $H'$ is one? (If the answer is “yes”, then the previous bullet point completely solves the problem for $n=8$, for example.)
If we take $u,v$ to be sufficiently general elements of the subfield of $k(x_4,y_4,\ldots,x_{n-1},y_{n-1})$ invariant under all permutations of $\{4,\ldots,n-1\}$ (i.e., generated by elementary symmetric functions of the $x$'s and those of the $y$'s), we get a construction invariant (precisely?) under $\mathfrak{S}_{n-4}$ seen inside $\mathfrak{S}_n$ as the pointwise stabilizer of $\{0,1,2,3\}$.
Failing a general characterization of symmetry subgroups of $n$-point constructions, I am particularly curious about the following case: if $\mathfrak{S}_5$ is included in $\mathfrak{S}_6$ as the stabilizer of a synthematic pentad (i.e., applying the outer automorphism of $\mathfrak{S}_6$ to the stabilizer of a point), does there exist construction on $6$ points having this particular symmetry group?
The motivation for this question arose when reading about Pascal's Hexagrammum Mysticum (see here about this) which involves a lot of constructions on a set of $6$ points with various symmetry groups inside $\mathfrak{S}_6$ (but by no means all: in particular, there is no point in the Hexagrammum Mysticum with the symmetry group given by the stabilizer of a pentad). Note that in the case of the Hexagrammum Mysticum the $6$ points are assumed to lie on a common conic, so this does not fall under the formalism of this question, but it is still related.
