Suppose that sequences $v_1,\ldots,v_m\in \{-1,+1\}^n$ and $u_1,\ldots,u_m\in \{-1,+1\}^n$ are isometric in $\mathbb{R}^n$, i.e. $u_j=Qv_j$ for some orthogonal matrix $Q$. What is the largest $m$ such that $Q$ is not an integer orthogonal matrix, i.e. one of the $2^n n!$ symmetries of the hypercube? In other words, what is the largest $m$ such that for some such isometric sequences $(u_j)$ and $(v_j)$ in $\{-1,+1\}^n$ there is no integer orthogonal matrix $Q$ such that $u_j=Qv_j$?
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5$\begingroup$ You might first ask what's the smallest $n$ for which this is possible at all. An upper bound is $16$, because there is more than one equivalence class of Hadamard matrices of order $16$. It seems a plausible guess that the optimum for all feasible $n$ (or maybe all sufficiently large $n$) will be obtained by starting from some small example, with parameters $(n_0, m_0)$, and forming a family of examples with parameters $(n, 2^{n-n_0} m_0)$ for each $n \geq n_0$ by extending $\{u_j\}$ and $\{v_j\}$ with all possible $\pm 1$ sequences of length $n-n_0$. $\endgroup$– Noam D. ElkiesCommented Dec 15, 2019 at 22:58
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1$\begingroup$ @NoamD.Elkies, I would accept your example as an answer, since this is more or less what I was looking for. $\endgroup$– D_809Commented Dec 16, 2019 at 0:34
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1$\begingroup$ Thanks; first let me think about this a bit more --- I think I can bring $n_0$ down to $7$ (which will also increase the largest possible $m$). $\endgroup$– Noam D. ElkiesCommented Dec 16, 2019 at 14:23
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