@DaveBenson One example is the subgroup of $\mathrm{Sym}(\aleph_\omega)$ of all permutations with support strictly less than $\aleph_\omega$. Is there a finitely generated example?

@MoisheKohan Interesting, thanks! Does this also rule out quasi-isometry with a 2-gen group?

@LSpice Yes. (Also, the phrase "Gelfand pair" was added after I made my comment.)

Replacing $x$ with $xy^{-1}$, this is equivalent to saying that $K$-biinvariance implies right invariance with respect to conjugates of $K$. There is a counterexample with $|G| = 6$.

Synonyms: regular uniform designs, configurations, tactical configurations, 1-designs. See en.wikipedia.org/wiki/…. These are also more-or-less equivalent to biregular bipartite graphs.

Won't this hold for any set with (nice boundary and) a point symmetry?

Not so fast. $w$ could be $x^k$ or $y^k$.

Any chance of getting further details about the lower bound? It seems surprising. Is it some sort of saddle-point analysis?

Hi, Padraig Ó Catháin and I were discussing this question and answer this week. This argument is very interesting. However, for any $(0,1)$-matrix one actually has the following variant of the Hadamard bound: $\det A \le (n+1)^{(n+1)/2} / 2^n$. This follows from $\det A = \det \begin{pmatrix} 1 & 1 \\ 0 & A\end{pmatrix} = 2^{-n} \det \begin{pmatrix} 1 & 1 \\ -1 & 2A - 1 \end{pmatrix}$ followed by the Hadamard bound for an $(n+1) \times (n+1)$ matrix. Therefore your upper bound $\lim f_{\mathrm{cir+}}$ can be improved to $0$.

Concretely, if $M$ is the Gram matrix of the symmetric bilinear form, the conformal orthogonal group is the group of matrices $g$ such that $g^T M g = \lambda M$ for some nonzero $\lambda \in K = \mathbf{F}_q$, and this defines a natural homomorphism $c : \mathrm{CO}(M) \to K^\times$, $g \mapsto \lambda$. Note that $\det(g)^2 = c(g)^{2n}$, so $\det(g) = \pm c(g)^n$, and $\mathrm{CSO}(M) = \{g \in \mathrm{CO}(M) : \det(g) = c(g)^n\}$.

Updated guess: CO is the group of similarities of the orthogonal form, i.e., elements that map the orthogonal form to a scalar multiple of itself. When $k$ is algebraically closed that is the same thing as scalars times O, but once you take $F$-fixed points it's not quite, but scalars times O is an index-two subgroup (for $q$ odd). Probably $\mathrm{CO}^\circ$ for Malle--Testerman corresponds to CSO for Magma. In which case, yes, $(\mathrm{PCO}^\circ_{2n})^+(q) = \mathtt{PCSOPLus}(2n, q)$.

@DaveBenson I must be wrong then. As far as I understood CSO means scalars x SO, so PCSO would be the same thing as PSO.

In characteristic 2 there is a further wrinkle to do with the fact that determinant is trivial in characteristic 2, but still $[\mathrm{CO}_{2n} : \mathrm{CO}_{2n}^\circ] = 2$. Use "pseudodeterminant" instead, and conventions vary about what $\mathrm{SO}$ means in char 2. I don't know what Magma believes, but maybe you can check by computing the order of $\mathrm{PSO}^+_8(2)$.

I think this is right, at least if $p$ is odd, though I'm not familiar with Magma. I believe $\mathrm{PCO}_{2n}^\circ$ is more or the less the same thing as $\mathrm{PSO}_{2n}$, though people who know more than me don't like to write the latter thing for some reason. So $(\mathrm{PCO}_{2n}^\circ)^+(q)$ is probably just $\mathrm{PSO}^+(2n, q)$ in Magma.