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Cosmas Zachos
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Even though the intro to your question conjures up Pauli's legendary quantization of the Hydrogen atom using the rotational so(3) symmetry and the suitably normalized LRL vector which can be combined with the above to yield the celebrated so(3)⊕so(3)~so(4) (note I am skipping groups in favor of Lie algebras, which are all that matter here!), your question is a thoroughly classical one.

Since I failed to understand your question (on account of the last two lines of it), I'll just jot down the straightforward part, the algebraic structure of the generators for the 3D canonical transformation generators, an sp(6), and their overlap with the above symmetry so(4).

For simplicity, I'll deconstruct, backtrack on the unravelling of the symmetry generators of the so(4). I'll consider the so(3) rotation subalgebra; and the three peculiar components of the LRL vector symmetry underlain by the astounding Lie scaling symmetry in which the coordinates r and the time t are scaled by different powers of a parameter λ,

$$ t \rightarrow \lambda^{3}t , \qquad \mathbf{r} \rightarrow \lambda^{2}\mathbf{r} , \qquad\mathbf{p} \rightarrow \frac{1}{\lambda}\mathbf{p}. $$ This transformation changes the total angular momentum L and energy E, $ L \rightarrow \lambda L, \qquad E \rightarrow \frac{1}{\lambda^{2}} E, $ but preserves their product $EL^2$. Therefore, the eccentricity e and the magnitude A are preserved... $$ A^2 = m^2 k^2 e^{2} = m^2 k^2 + 2 m E L^2.$$ The direction of A is preserved as well, since the semiaxes are not altered by a global scaling. This transformation also preserves Kepler's third law,...

Now you may easily convince yourself that rotations (of r and p in the same sense!) preserve Poisson brackets, so provide a canonical transformation; but, visibly from the above, the Lie tfmations scale PBs by λ, so they are not canonical transformations. So these three transformations are not canonical, indicating, straightforwardly, that only rotations are canonical transformations, so they overlap with only 3 of the 21 generators of the sp(6), as indicated in your formula (as I imagined it corrected; are you faked out by sp(4)~so(5)?).

I have a sense that, by failing to provide a reference, you have withheld relevant context.

Cosmas Zachos
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