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    $\begingroup$ How is an 8 by 8 Hadamard matrix "not orthogonal in character"? $\endgroup$ Commented Feb 2, 2020 at 20:56
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    $\begingroup$ Echoing @user44191: Hadamard matrices are precisely the $\pm 1$-valued matrices such that $A^\top A$ is equal to a scalar multiple of the identity, so rescaling one of these always gives an orthogonal matrix whose entries share the same absolute value. The determinant 1 condition should be easy to check in various examples. $\endgroup$ Commented Feb 2, 2020 at 22:10
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    $\begingroup$ Also, there are matrices called complex Hadamard matrices of all orders which can supply such orthogonal matrices with entries from the complex numbers. Gerhard "Going Off In Another Dimension" Paseman, 2020.02.02. $\endgroup$ Commented Feb 2, 2020 at 22:36
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    $\begingroup$ Thanks for the comments! Well, I had taken the Kronecker product of a $2 \times 2$ and a $4 \times 4$ Hadamard matrix to presumably get an $8 \times 8$ one, which proved to be not orthogonal, that is its matrix product with its transpose was not proportional to the identity. But it looks like I should probably recheck my computations/references--in view of the comments---just rechecked. My $H_2$ was miscoded--so mea culpa and the $H_8$ is orthogonal. But what about $n$ not a mulitple of 4. $\endgroup$ Commented Feb 2, 2020 at 23:45
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    $\begingroup$ Note that the construction works both ways; the existence of matrices of the form you're looking for is precisely equivalent to the existence of a Hadamard matrix, by scaling of the entries. The Wikipedia page on Hadamard matrices notes that they can only exist in dimensions $1$, $2$, and $4n$, so you won't find any other examples. $\endgroup$ Commented Feb 3, 2020 at 0:29