Let $m\geq 3$ be fixed and $n\to\infty$. Consider $v=(v_j)_{j\leq m}$ with $v_1,\ldots,v_m\in \{-1,+1\}^n$. Let:

- $N_I(v)$ be the number of sequences $u_1,\ldots,u_m\in \{-1,+1\}^n$ isometric to $v$ in $\mathbb{R}^n$, i.e. $u_j=Qv_j$ for some orthogonal matrix $Q$.
- $N_S(v)$ be the number of sequences $u_1,\ldots,u_m\in \{-1,+1\}^n$ isometric to $v$ in $\{-1,+1\}^n$, i.e. $u_j=Qv_j$ for some
**integer**orthogonal matrix $Q$.

How does the average of $N_S(v)/N_I(v)$ over $2^{nm}$ choices of the sequence $v$ behave as $n\to\infty$. In other words, how typical are hypercube symmetries among all isometries of finite subsets of $\{-1,+1\}^n$?

Instead of looking at the average of the ratio, we can compute the corresponding probabilities. Let $u_1,\ldots,u_m$ and $v_1,\ldots, v_m$ be chosen uniformly at random from $\{-1,+1\}^n$ and let $$U=(u_i\cdot u_j)_{1\leq i,j\leq m},\quad V=(v_i\cdot v_j)_{1\leq i,j\leq m}$$ be the Gram matrices for these two sequences, so the probability that these two sequences are isometric in $\mathbb{R}^n$ is $\Pr(U=V).$ It is not difficult to estimate using Sterling's formula that the probability that these sequences are isometric in $\{-1,+1\}^n$ has the leading order $$\sim\left(\frac{1}{\sqrt{n}}\right)^{2^m-1}.$$ The probability $\Pr(U=V)$ is bigger, so it is reasonable to expect that the leading order $$\Pr(U=V)\sim \left(\frac{1}{\sqrt{n}}\right)^{c_m}$$ for some constant $c_m\leq 2^m-1$. Is $c_m<2^m-1$? Any pointers are appreciated.