Is the nonnegative orthant isometric to itself under orthogonal mapping?

Problem description:

(Informal). I simply want to know if there exists a necessary and (or at least) sufficient condition for an orthogonal matrix to map every point of the nonnegative orthant (say on $\mathbb{R}^m$) to itself.

(Formal). Let $\mathscr{D}_m = \{(x_1,\ldots,x_m)\in\mathbb{R}^m\,:\, x_i\geq 0,\ i=1,2,\ldots,m\}$. For a given $\mathbf{Q}$ an $m\times m$ real orthogonal matrix define the image $\mathscr{D}_m^{*} = \{\mathbf{y}\in\mathbb{R}^m\,:\, \mathbf{y}=\mathbf{Q}\mathbf{x},\, \mathbf{x}\in\mathscr{D}\}$. Is there a necessary and (or at least) sufficient condition on $\mathbf{Q}$ in order for $\mathscr{D}_m = \mathscr{D}^{*}_m$?

Motivation/Context of the problem: (Change-of-variable in Integration)

Let $\mathbf{S}$ be an $m\times m$ positive definite matrix. It is well-known that there exists an orthogonal matrix $\mathbf{Q}$ and a diagonal matrix $\boldsymbol{\Lambda}$ such that $\mathbf{Q}^{\top}\mathbf{S}\mathbf{Q} = \boldsymbol{\Lambda}$. With this, we have the following equality of $m$-variate integration via the classic (multivariate) change-of-variable, noting that the absolute value of the Jacobian is $|\text{det}(\mathbf{Q})| = 1$, $$\int_{\mathscr{D}_m} \exp\{\mathbf{x}^{\top}\mathbf{S}\,\mathbf{x}\}\cdot\exp\{-\frac{1}{2}\mathbf{x}^{\top}\mathbf{x}\}\,\text{d}\mathbf{x} = \int_{\mathscr{D}^{*}_m} \exp\{\mathbf{y}^{\top}\boldsymbol{\Lambda}\,\mathbf{y}\}\cdot\exp\{-\frac{1}{2}\mathbf{y}^{\top}\mathbf{y}\}\,\text{d}\mathbf{y}$$ where $\mathscr{D}_m$ and $\mathscr{D}^{*}_m$ are given in the (formal) problem description. The left-hand-side can be calculated easily if $\mathscr{D}_m = \mathscr{D}^{*}_m$ which gives our motivation of the problem stated above.

Attempt:

Honestly speaking, I failed to find from the web for a simpler explanation. I came across with the article of Barker and Foran Self-Dual Cones in Euclidean Spaces with an abstract that contains the following line..

...We begin by giving necessary and sufficient for a cone to be the orthogonal transform of the positive orthant...''

I humbly accept that I am new to this topic. It is even my first time to hear that the nonnegative orthant is a symmetric self-dual cone. I have basic Linear Algebra and Calculus as background. Hope anyone can help me with this.

EDIT: I found this link: Find a nonnegative basis of a matrix nullspace / kernel that relates to this problem.

• There is something on this problem in the Perron-Frobenius theorem, do you know it? I am not too sure about this. – Giuseppe Negro Sep 11 '18 at 12:03
• If the rows of an orthogonal matrix are non negative then the matrix is the identity. I suspect that no matrix except the identity fixes the positive orthant. – Giuseppe Negro Sep 11 '18 at 14:08
• @mr_e_man: Right, thank you. The corrected statement is: "every orthogonal matrix with non-negative entries is a permutation matrix", and the corrected conjecture is that only the only orthogonal matrices that fix $\mathscr D$ are permutations. – Giuseppe Negro Sep 11 '18 at 16:02
• In 3D, it's obvious that the positive octant (a "corner of a cube") has exactly 6 symmetries: 3 reflections, 2 rotations (120deg), and the identity. I believe this generalizes to nD, with all symmetries being compositions of reflections that swap 2 axes. – mr_e_man Sep 11 '18 at 16:03
• I don't know how to formalize this, but the positive orthant is the convex hull of the positive coordinate axes, and the axes are the "edges" of the orthant, so they should have the same symmetries. A symmetry of the axes must send every axis to another axis, and must be invertible; these are indeed permutation matrices. – mr_e_man Sep 11 '18 at 16:17