@DaveBenson I should have expanded this: that $\Gamma$ contains a subgroup isomorphic to $\Gamma/B$ which is a complement to $B$.

Thank you @DerekHolt - I've made the edits you suggested. Do you have a reference for Q1 and Q2? If so, I'll be happy to accept this as an answer. And I agree that your counterexample shows that the general situation is more complicated than I had hoped.

I don't know a name for groups with this property, but such questions are studied with the transfer homomorphism and the focal subgroup theorem.

The Paley II construction gives a symmetric Hadamard matrix of order $2q+2$ for any prime power $q\equiv 1 \mod 4$, with trace $0$. A symmetric Bush-type Hadamard matrix can be constructed with order $16n^2$ whenever there is a symmetric Hadamard matrix of order $4n$, these always have trace $16n^2$. It's conjectured that a symmetric Hadamard matrix exists at all orders $4n$ and that a Bush-type matrix exists for all $4n^2$.

Ah - that's true. I'm not aware of any literature about this equivalence operation on rectangular matrices. Are you aware of any invariants of $X$ that are preserved?

The reduced row echelon form is unique, so you may as well start there. You can apply an arbitrary permutation of rows, apply the inverse permutation on the first block of columns to recover the identity matrix, and then apply an arbitrary permutation to the remaining columns. So I guess that $X$ is determined up to arbitrary permutation of rows and columns. Deciding whether given $X_{1}$ and $X_{2}$ are equivalent under this operation is more-or-less isomorphism of bipartite graphs.

Up to renormalising and reordering rows/columns you have the direct sum of two Hadamard matrices. It seems likely that the stabiliser of this matrix in the monomial group would respect this direct sum structure, and the automorphism group of each summand (as a Hadamard matrix) projects onto $S_4$. So it's likely that the permutation quotient is a wreath product $S_4 \wr S_2$ (which is maximal in $S_8$). It's known that no larger Hadamard matrix has such a highly transitive automorphism group.

Granville and Martin wrote a paper where this is discussed: arxiv.org/pdf/math/0408319.pdf They explain many details of the behaviour of these 'prime races'. Some buckets are fuller than others most of the time, as you have observed. They can quantify by how much some buckets lead, and how often. The tie all this to the Riemann hypothesis - it's a really nice paper.

The preprint mentioned by Martin has a flaw - some of the claims about cyclotomic fields work only for $\mathbb{Q}[\zeta_{2^{k}}]$ and results of Turyn rule out further circulant Hadamard matrices for those orders.

@BCS Sequences in which all non-trivial autocorrelations are -1 can be used to construct Hadamard matrices with circulant core. The Paley and Sylvester matrices arise in this way from Legendre sequences constructed from quadratic residues and m-sequences respectively (as in the answer above). You just add a row and column of 1's. Otherwise, Hadamard matrices are not in general of use. The circulant of order 4 has been noted, Bernhard Schmidt has shown there are no others of order < 10^20 or so.

@FrancoisZiegler Thank you - this can't be the first place the idea appeared, but nice to see it written down, in whatever format :)

If $n+1$ is the order of a Hadamard matrix, then normalising and removing the row and column of all-ones gives a matrix such that $MM^{\top} = (n+1)I - J$. Taking $A, B, C$ all equal to $M$ gives a solution. For question (2) the techniques used for finding Williamson matrices could probably be adapted to find solutions computationally.