Sorry. I just noticed that my comment about the count of orbits is directly addressed (for $d=2$, anyway) in a "Related" MO problem to the left: mathoverflow.net/questions/41337/…

I like your canonical form and Brendan's answer below. I just wanted to comment that Burnside's lemma says there are \begin{equation} \frac{1}{|S_N|} \sum_{\alpha \in S_N} |\text{centralizer}(\alpha)|^d = \frac{1}{N!} \sum_{\alpha} \left( \sum_{\chi} \chi^2(\alpha) \right)^d \end{equation} different orbits of $G^d$ under simultaneous conjugation. So you can compute the corresponding count of types rather easily (but it involves the character table).

I may not understand properly, but doesn't $c=1$ work? The goal is to get $c$ as small as possible. Two observations: (1) From simple counting, it's clear that we need $c > 1/3$. (2) If you specify your set $P \subset [v]$ ahead of time and then quantify over all STS on $[v]$, you run into counterexamples for $c$ below $1/2$.

OK good. I am outside my comfort level here, but I agree this should be well known. Do you vaguely recall if the determinant was of some Hessenberg matrix?

What we could really use is, for each square, a list of multiples of that square which are mutually disjoint. With enough such examples, we can recursively split up squares into smaller and smaller squares and guarantee they are all disjoint.

It just occurred to me that our 25 squares don't need to be of the same order (number of constituent squares) -- just the same size.

Flats are also known as subdesigns, or subspaces. By the way, I am interested in designs such that any "small" collection of points is contained in a proper flat/subdesign. They exist with surprising abundance, not just in affine or projective space of high dimension. There are lots of tangents in these cases.

It is conjectured, I think, that the number of 1-factorizations of $K_n$ is about $(n/e^2)^{n^2/2}$. Bounds are known, but where the base of the exponent differs by a constant multiple. We want 1-factorizations of $K_{n,n}$ instead. Do the techniques extend somehow? With not much effort we get something in between $(C_1 n)^{n^2/2}$ and $(C_2 n)^{2n^2}$, but there is a huge gap.

I guess a line could be defined as the locus of points for which one has equality in the triangle inequality with respect to two given points. Then, Masked Avenger's comment gives the parabola. I would be interested to see this parabola for line $\langle (-1,2) \rangle$ and focus $(2,1)$.

You probably know this already, but $S$ has your Ore property whenever $S$ is contained in a proper flat, and also whenever $P$ is large relative to (quadratic in?) $S$.

As discussed after my incomplete solution: The answer is yes if we have 25 mutually disjoint perfect squared squares of some order. This page discusses disjoint families: squaring.net/sq/sr/spsr/spsr_dnt.html. All indications are that a family of 25 should exist. But, unfortunately, I see no elementary way to get the family (although at first I thought it should be easy via compound squares).

Yes... I am concerned too. I am thinking about a fix to easily get the 25 squares.