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This is a cross post from MSE, as it seems the partial answer I got then was deleted, so I ask it again here.

Let $\prod_{i\in I}p_{i}^{a_{i}}$ be the prime factorization of a positive integer $n$ and let's consider the $\Omega(n)$-parallelotope built with, for all $i\in I$, $a_{i}$ pairwise orthogonal vectors of norm $p_{i}$, so that there are $\omega(n)$ different lengths for the sides of this parallelotope. Let's consider the isometry group of this parallelotope, as a part of the affine space $\mathbb{R}^{\Omega(n)}$, denoted by $G(n)$. Of course, any permutation of $ I $ gives rise to the same group, and as such should be seen as an automorphism thereof.

Can this group provide information of arithmetical interest about $n$? I'm especially interested in estimating the size of $ \pi_{G}(x) : =\{n\leq x, G(n)\cong G\} $.

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  • $\begingroup$ Do you mean the direct sum of hyperoctahedral groups, $\bigoplus_{i \in I} B_{a_i}$? $\endgroup$
    – Adam P. Goucher
    Commented Mar 5, 2018 at 0:16
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    $\begingroup$ Obviously $G(n)$ contains an appropriate group of coordinate-permutation isometries; for example, if $n=4m$ where $m$ is odd and squarefree, then $G(n)$ includes the involution that exchanges the two coordinates where $a_i$ has norm $2$ and fixes the other coordinates. Do you have an example where $G(n)$ is strictly larger than this trivial subgroup? If there are no such examples, then you're just asking about the counting function of integers with a fixed factorization type, which is an easy question. $\endgroup$
    – Greg Martin
    Commented Mar 5, 2018 at 1:27
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    $\begingroup$ @Greg Martin : it seems to me that the involution you consider is geometrically the reflexion with respect to a hyperplane that contains the diagonal of the square with size 2, so yes there are other elements : in your example the dihedral group of order 8 is a subgroup of $ G(n) $ . $\endgroup$
    – Sylvain JULIEN
    Commented Mar 5, 2018 at 6:28
  • $\begingroup$ Great point, showing I need to recalibrate. Ok, if $n=\prod_{i=1}^k p_i^{a_i}$, then the parallelotope is the direct product of $k$ hypercubes, where the $i$th hypercube has dimension $a_i$. Therefore $G(n)$ contains the product of the isometry groups of these hypercubes. (Even a $1$-dimensional hypercube has a nontrivial isometry, I should have realized.) Again the question to consider is: are there ever any additional isometries? for if not, it again boils down to integers with a prescribed factorization type. $\endgroup$
    – Greg Martin
    Commented Mar 5, 2018 at 8:10
  • $\begingroup$ I guess you're right but still, it may give rise to interesting questions like détermining the sets $ \mathbb{N}_{G} : =\{n,G(n)\cong G\} $ that form a partition of $ N $ or the sequence of '$ G $ -type radii of $ n $' defined as the non negative integers $ r $ such that $ G(n-r)\cong G(n+r)\cong G $ . Of course $ \mathbb{N}_{\mathbb{Z}/2\mathbb{Z}} $ is the set of prime numbers. $\endgroup$
    – Sylvain JULIEN
    Commented Mar 5, 2018 at 8:50

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