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Well, if $Q\in\mathrm{SO}(n)$ satisfies your (original) conditions then so does $ \frac1{\sqrt2} \begin{pmatrix} -{}^tQ&Q\\ \phantom{-}{}^tQ&Q \end{pmatrix} \in\mathrm{SO}(2n). $

Added: As S. Stadnicki since commented, your desired set $\mathrm S$ of possible orders $n$ is contained in and conjecturally equal to $\{1,2\}\cup4\mathbf N$ (Hadamard conjecture); the above construction ($\cong$ Sylvester’s) just shows $\smash{2^{\mathbf N}\subset\mathrm S}$. (Voting to close, as R. Israel had really said all this at the mis-linked question.)

Well, if $Q\in\mathrm{SO}(n)$ satisfies your (original) conditions then so does $ \frac1{\sqrt2} \begin{pmatrix} -{}^tQ&Q\\ \phantom{-}{}^tQ&Q \end{pmatrix} \in\mathrm{SO}(2n). $

Added: As S. Stadnicki since commented, your desired set $\mathrm S$ of possible orders $n$ is contained in and conjecturally equal to $\{1,2\}\cup4\mathbf N$ (Hadamard conjecture);* the above construction ($\cong$ Sylvester’s) just shows $\smash{2^{\mathbf N}\subset\mathrm S}$. (Voting to close, as R. Israel had really said all this at the mis-linked question.)

* Paley (1933, front page) proved $\mathrm S\subset\{1,2\}\cup4\mathbf N$ thus: Assume w.l.o.g. that all entries are $\pm1$ and $Q$ has 3 distinct columns $u,v,w$. Then their orthogonality gives $$ n=\|u\|^2=\langle u+v,u+w\rangle=\sum\nolimits_i(u_i+v_i)(u_i+w_i), $$ a sum all of whose terms are $0$ or $\pm 4$.

Well, if $Q\in\mathrm{SO}(n)$ satisfies your (original) conditions then so does $ \frac1{\sqrt2} \begin{pmatrix} -{}^tQ&Q\\ \phantom{-}{}^tQ&Q \end{pmatrix} \in\mathrm{SO}(2n). $

Added: As S. Stadnicki since commented, your desired set $\mathrm S$ of possible orders $n$ is contained in and conjecturally equal to $4\mathbf N$ (Hadamard conjecture); the above construction (essentially Sylvester’s) just shows $\smash{2^{\mathbf N}\subset\mathrm S}$. (Voting to close, as R. Israel had already said all this at the mis-linked question.)

Well, if $Q\in\mathrm{SO}(n)$ satisfies your (original) conditions then so does $ \frac1{\sqrt2} \begin{pmatrix} -{}^tQ&Q\\ \phantom{-}{}^tQ&Q \end{pmatrix} \in\mathrm{SO}(2n). $

Added: As S. Stadnicki since commented, your desired set $\mathrm S$ of possible orders $n$ is contained in and conjecturally equal to $\{1,2\}\cup4\mathbf N$ (Hadamard conjecture); the above construction ($\cong$ Sylvester’s) just shows $\smash{2^{\mathbf N}\subset\mathrm S}$. (Voting to close, as R. Israel had really said all this at the mis-linked question.)

Well, if $Q\in\mathrm{SO}(n)$ satisfies your (original) conditions then so does $ \frac1{\sqrt2} \begin{pmatrix} -{}^tQ&Q\\ \phantom{-}{}^tQ&Q \end{pmatrix} \in\mathrm{SO}(2n). $

Added: As S. Stadnicki since commented, your desired set $\mathrm S$ of possible orders $n$ is contained in and conjecturally equal to $4\mathbf N$ (Hadamard conjecture); the above construction (essentially Sylvester’s) just shows $\smash{2^{\mathbf N}\subset\mathrm S}$.

Well, if $Q\in\mathrm{SO}(n)$ satisfies your (original) conditions then so does $ \frac1{\sqrt2} \begin{pmatrix} -{}^tQ&Q\\ \phantom{-}{}^tQ&Q \end{pmatrix} \in\mathrm{SO}(2n). $

Added: As S. Stadnicki since commented, your desired set $\mathrm S$ of possible orders $n$ is contained in and conjecturally equal to $4\mathbf N$ (Hadamard conjecture); the above construction (essentially Sylvester’s) just shows $\smash{2^{\mathbf N}\subset\mathrm S}$. (Voting to close, as R. Israel had already said all this at the mis-linked question.)