I'm not sure I fully understand the question so this is more of a query than anything - the eigenvalues of $a_i \otimes I_{D/d}$ are just the eigenvalues of $a_i$ repeated $D/d$ times so it seems that if the eigenvalues of any of the $A_i$ are distinct then $d = D$. On the other hand to make $d$ any smaller than that you need the eigenvalues of $A_i$ to share multiplicities somehow?

This paper seems relevant: arxiv.org/abs/2203.16439, they have a github repository here github.com/Qomo-CHENG/Hadamard_bent

@KajalDas The Fourier series gives you an isometry from $L^2([0,1])$ to $l^2$ (the space of square summable sequences. What I was suggesting was to construct a unitary on $L^2([0,1])$ by mapping to $l^2$, then applying some simple unitary on $l^2$ (for example a diagonal one, which just multiplies each component by a complex number of absolute value $a$) then applying the inverse Fourier transform to this new $l^2$ sequence which gives you a new $L^2([0,1])$ function. The Riesz–Fischer theorem ensures this process gives you a unitary on $L^2([0,1])$.

Is it possible to be more specific about what you're looking for? There are a lot of examples you can cook up. One easy approach if your space is $L^2([0,1])$ is to mess with the Fourier series, e.g. Fourier transform your vector, change the Fourier series somehow (apply a permutation or stick some phases on some of the components or something) and Fourier transform back.

@HaukeReddmann yes - this diagonalization is exactly what you want in this case because your form is given by $x^T A y$, and not $x^\dagger A y$.

@MichaelAlbanese Mea culpa, I missed the restriction that the matrices are integer valued.

@AlexM. I am not satisfied with $\ln{\sqrt[4]{2}}$, unless it is the case that every $\mathrm{SO}(2n, \mathbb{R})$ is the exponential of an $\mathfrak{so}(2n,\mathbb{R})$ matrix satisfying that bound, which I think is not true (or unless a counterexample exists showing that this bound is necessary to ensure convergence in this case).

@AlexM. That I've read the wiki page and that article recently. The condition given is not necessary for convergence, and I was hoping that a much weaker condition suffices for matrices in the special orthogonal Lie algebra. The couterexamples given in the wiki page and the article are such that there does not exist a matrix in the Lie algebra that exponentiates to the product of the two exponentials. This case never happens for the special orthogonal group/algebra since the exponential is surjective. I don't know if there are cases of non-convergence that are not of this form.