Suppose $\vec x$ is a (non-zero) vector with integer coordinates in $\mathbb R^n$ such that $\|\vec x\| \in \mathbb Z$. Is it true that there is an orthogonal basis of $\mathbb R^n$ containing $\vec x$, consisting of vectors with integer coordinates, all with the same length?
For example: let $\vec x = \left<2,10,11\right>$, so $\|\vec x\| = 15$. Then the vectors $\left<14,-5,2\right>$ and $\left<5,10,-10\right>$ complete a basis of $\mathbb R^3$.
I've checked all such integer vectors in $\mathbb R^3$ with an integer length up to 17 and found no counter examples. Moreover, these can always be arranged as a symmetric matrix, possibly changing the order (or permuting the coordinates) and changing signs (edit: not always; see answer below). For example, the vectors above can be arranged as:
This is easily true if $n$ is even (edit: rather, if $n=4$), you can simply permute the entries of $\vec x$ (altering signs appropriately) to find $n-1$ other vectors orthogonal to $\vec x$. In $\mathbb R^3$, finding a second integer vector is sufficient because the cross product of the two (divided by $\|\vec x\|$) will give the third vector of such a basis. Then it is straightforward to prove for special cases (like $\vec x = \left<1,2m,2m^2 \right>$, $m\in \mathbb Z$), but I can't think of a good reason why this should be true in general.
Edited, hopefully for clarity, by o.p. The original phrasing and title referred to $\mathbb Z^n$.