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,
By suitably normalizing the LRL vector to ${\mathbf D}\equiv {\mathbf A}/\sqrt{-2H}$ and unravelling to chiral components, A = L + D , B = L - D , you can show the A s close into an su(2) commuting with the su(2) of the B s. So, then, a total so(4). This, now untwisted, so(4) is a bona-fide symmetry and entirely canonical, and a subgroup of sp(6). That is, for the six parameters a, b, $$\delta {\mathbf q} = \{ {\mathbf q} , {\mathbf a\cdot A}\},\qquad \delta {\mathbf p} = \{ {\mathbf p} , {\mathbf a\cdot A}\} \\ \delta' {\mathbf q} = \{ {\mathbf q} , {\mathbf b\cdot B}\},\qquad \delta '{\mathbf p} = \{ {\mathbf p} , {\mathbf b\cdot B}\} $$ preserve the Poisson brackets, as you may (should) demonstrate by use. (For example, as per your request in the comment, the first order variation of $\{ q^i,p^j \}=\delta^{ij}$ vanishes, $$ \{ \delta q^i,p^j \} + \{ q^i,\delta p^j \} = \{ \{ q^i, {\mathbf a\cdot A}\} \},p^j \} + \{ q^i, \{ p^j,{\mathbf a\cdot A}\} \} \} \\ =-\{ \delta^{ij} , {\mathbf a\cdot A}\} - \{ \{ {\mathbf a\cdot A}\},p^j \}, q^i \} + \{ q^i, \{ p^j,{\mathbf a\cdot A}\} \} \}=0 $$ by the Jacobi identity, and likewise for all other PBs and generators.)
It hardly matters, since Pauli's algebraic maneuver, upon quantization, works just fine, anyway. One never really had to think about canonical transformations, and how motion is a canonical transformation with time as the infinitesimal parameter, and the whole shebang...