<s>[INST] A certain unique rotation matrix One can find that the matrix
$A=\begin{bmatrix}
    -\dfrac{1}{3} &     \dfrac{2}{3}      &     \dfrac{2}{3}   \\
     \dfrac{2}{3}  &    -\dfrac{1}{3}   &   \dfrac{2}{3}    \\
    \dfrac{2}{3} &    \dfrac{2}{3}  &    -\dfrac{1}{3}  \\
        \end{bmatrix}  $  
is at the same time $3D$ rotation  matrix and for it the sum of  entries  in  every  column (row)  is constant (here $-\dfrac{1}{3}+   \dfrac{2}{3}     +     \dfrac{2}{3} = 1)$.
 The same is true if we change  the order of columns in it.
For example:
$A_1=\begin{bmatrix}
    \dfrac{2}{3} &     \dfrac{2}{3}      &     -\dfrac{1}{3}   \\
     -\dfrac{1}{3}  &    \dfrac{2}{3}   &   \dfrac{2}{3}    \\
    \dfrac{2}{3} &    -\dfrac{1}{3}  &    \dfrac{2}{3}  \\
        \end{bmatrix}  $       $A_2=\begin{bmatrix}
     \dfrac{2}{3} &     -\dfrac{1}{3}      &     \dfrac{2}{3}   \\
     \dfrac{2}{3}  &     \dfrac{2}{3}   &   -\dfrac{1}{3}    \\
    -\dfrac{1}{3} &    \dfrac{2}{3}  &     \dfrac{2}{3}  \\
        \end{bmatrix}  $   
Questions:
Is it any systematic way to find other non-trivial (without $0$ and $1$) rotation matrices with this property?
Especially it is interesting whether the above rotation matrices are the only ones with rational entries ?- maybe someone knows other rotation matrices exist where the sum of entries is constant..
and...
Can it be proved in some way that if the sum of entries in columns for a  rotation matrix is constant then it should be equal to the length of  column vectors?
 [/INST] This answer is not an independent one, it's just a supplement to Yves'
answer to have made his formula for generating rational values of required matrix more explicit (what is important for me and maybe useful for others)
So we have Yves' equation  $ p^2+3g^2=1$ what can be written as $\left({\dfrac{a}{d}}\right)^2+3{\left({\dfrac{b}{d}}\right)^2} =1 $   
then    $ a^2+3b^2=d^2 $
 and $(d-a)(d+a)=3b^2$  so we have     $ d-a=3 $ and $d+a=b^2$.
 Solution for $a$, $d$ as a  $f(b)$: $d=\dfrac{b^2+3 }{2}$ 
$a=\dfrac{b^2-3 }{2}$.  
Now we have generator for numbers $a$ and $d$ assuming $b$ is odd.   
Examples (when $b$ is coprime with 3, although it seems not to be neccesary at general case): 
$b=5$ : $d=\dfrac{5^2+3 }{2}=14$   $a=\dfrac{5^2-3 }{2}=11$.
$b=7$ : $d=\dfrac{7^2+3 } {2}=26$  $a=\dfrac{7^2-3}{2}=23$.
$b=11$ : $d=\dfrac{11^2+3 }{2}=62 $   $a=\dfrac{11^2-3}{2}=59$.
$b=13$ : $d=\dfrac{13^2+3 }{2}=86 $   $a=\dfrac{13^2-3}{2}=83$. ...
...infinite number of solutions..
Although there is an infinite number of matrices with properties given in the question the matrix listed as first   seems still to be unique: it is  the sole non-trivial descendant of $4$ (!) families of matrices: 


*

*matrices with rational entries,

*rotation matrices,   

*matrices with constant row (column) sum of   entries, 

*and  symmetrical ones.

</s>