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:
$$\begin{bmatrix}14&-5&2\\-5&-10&10\\2&10&11\end{bmatrix}$$
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 $\mathbb Z^n$.
Well, if your real question is what @user1551 says, then it is probably true in $\mathbb Z^3.$ See the paper I call Pall_Automorphs_1940 at one of my websites, TERNARY. All that is necessary for this to settle dimension 3 is this: If $n$ is odd and $$ n^2 = x^2 + y^2 + z^2 $$ with $x$ odd, then we can find $a,b,c,d$ such that $$ x = a^2 + b^2 - c^2 - d^2, \; \; y = 2(ad - bc), \; \; z = 2(ac+bd) $$ and $$ n = a^2 + b^2 + c^2 + d^2. $$ This seems likely to me, and may be in one of Pall's articles. He and Jones found every way to use integral quaternions in studying the sum of three squares.
Other than that, I will need to think about it.
The shortest integral length for which a 3 by 3 such matrix cannot be made symmetric is 39, as in $$ \left( \begin{array}{ccc} 29 & 22 & 14 \\ 2 & 19 & -34 \\ -26 & 26 & 13 \end{array} \right) . $$ As the only repeated absolute value is 26, you do not have the three pairs of repeated absolute values necessary for transpositions of rows, or columns, and negation of either to result in symmetry. Actually, I suppose there could be some shorter length where some combinatoric problem prevents symmetry. Hard to tell.
The shortest integral length where you get nine distinct absolute values is 57, as in
$$ \left( \begin{array}{ccc} 47 & 28 & 16 \\ 4 & 23 & -52 \\ -32 & 44 & 17 \end{array} \right) . $$
The dimension 3 case is looking very good. I was looking in Dickson's History, volume II, page 271, where he mentions briefly that H. Schubert suggested this in 1902: we consider $4x^2 + 4y^2 + z^2 = n^2 $ with $z,n$ odd and $\gcd(n,z) = 1.$ This is not every case but it is a good start. It follows that with $$ u = \frac{1}{2}(n-z), \; \; v = \frac{1}{2}(n+z), $$ we also get $\gcd(n,z) = 1.$ Then, from $$ uv = x^2 + y^2, $$ we know that the two factors are separately sums of two squares, that is $$ v = a^2 + b^2, \; \; u = c^2 + d^2 $$ for some $a,b,c,d.$ But this immediately gives $$ n = a^2 + b^2 + c^2 + d^2, \; \; z = a^2 + b^2 - c^2 - d^2. $$ That is most of the battle.
FOUND IT. For dimension 3, this is Theorem 3 on page 176 of Jones and Pall (1939) which is on that website, taking $\lambda = 1.$ So dimension 3 is affirmative.
Dimension 4 is also affirmative, because you may begin with any quaternion $t$ and make the evident matrix out of $t,it,jt,kt.$
However, I do not see why that says anything about dimension 6. Given a row of six integers, certainly a rearrangement of pairs, with one extra minus sign for each pair, gives an orthogonal row. I don't see what to do after that. It is always possible that the octonions give some method for dimensions 5,6,7,8, but I would not be too sure about dimension 9 and above.