Are there strategies for identifying hidden symmetries numerically? I have a quantum mechanical system with a Hamiltonian, $H$, that has degeneracy in its spectrum. I have identified a matrix that commutes with this Hamiltonian, let's call it $B$, so $[B,H] = 0$. This matrix represents $SO(3)$ rotations of the system about the $\hat{z}$ axis. I observe that in the basis where $H$ is diagonal, $B$ takes a block diagonal form, where the blocks of B are in direct correspondence with the degenerate eigenspaces in the diagonal $H$. Now I look further at these blocks of $B$, for example I select out the first block, which happens to be 8-dimensional (I could note that my full system happens to be 27-dimensional, but I don't think that's a crucial detail to my question). I can diagonalize this block, and I find the following diagonal form,
\begin{align}
    \begin{pmatrix}
        2 & 0  & 0 & 0  & 0  & 0 & 0 & 0 \\
        0 & -2 & 0 & 0  & 0  & 0 & 0 & 0 \\
        0 &  0 & 1 & 0  & 0  & 0 & 0 & 0 \\
        0 &  0 & 0 & -1 & 0  & 0 & 0 & 0 \\
        0 &  0 & 0 & 0  & -1 & 0 & 0 & 0 \\
        0 &  0 & 0 & 0  & 0  & 1 & 0 & 0 \\
        0 &  0 & 0 & 0  & 0  & 0 & 0 & 0 \\
        0 &  0 & 0 & 0  & 0  & 0 & 0 & 0 
    \end{pmatrix}
\end{align}
From this I understand that this block was a reducible representation of $\mathfrak{so}(3)$, which breaks down into irreducible representations of $\mathfrak{so}(3)$, since I recognize these as the eigenvalues of the $L_3$  operator, for $j = 2$ and $j=1$. The fact that I'm observing a degeneracy in $H$'s spectrum, where a single degenerate eigenspace transforms under a reducible representation representation of $\mathfrak{so}(3)$ (constructed from these $j=2$ and $j=1$ irred. reps.) tells me that the symmetry group of $H$ must be larger than $SO(3)$, so as to explain this accidental degeneracy. I'm saying there's some hidden symmetry here.
This seems very reminiscent of the situation with the Hydrogen atom, where the full group of the Hamiltonian is $SO(4)$. My question though is if there are any strategies I could employ for identifying the larger symmetry group of my $H$, just from having the matrix, and thus being able to compute it's eigenvectors and eigenvalues numerically as I have done?
(I'm not the most confident with talking about representation theory and how it relates to degeneracy, so if I've written something that seems suspect or inconsistent, please let me know.)
Edit: In response to comments, I'm providing the $H$ and $B$ matrices in the basis where $H$ is diagonal, which I referred to above,
First is $H_D$, the $D$ subscript denotes this basis where $H$ is diagonal,
\begin{align}
H_D = \text{diag}(-2, -2, -2, -2, -2, -2, -2, -2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 10, 10, 10)
\end{align}
and the following matrix is $B_D$, where the $D$ subscript again denotes that were in this special basis, and I have further diagonalized all of the blocks,
\begin{align}
B_D = \text{diag}(2, -2, 1, -1, -1, 1, 0, 0, -3, 3, -2, 2, -2, -1, 2, 1, 1, -1, 1, -1, 0, 0, 0, 0, 1, -1, 0)
\end{align}
 A: Not an answer, just a long note to avoid sequential remarks/comments.
Firstly, you may interchange entries in your two 27-vectors simultaneously, by suitable permutation/interchange operators; so, within an energy level, you may rearrange the B entries to make them look more ordered. I'll do that with your energy eigenvalues, leaving the null ones, for the end.
Your 27 reduces to

*

*An octet 8 with $E=-2$ and $j=2$ and $j=1$;


*A  singlet 1 with E=4 and  $j=0$;


*A triplet 3 with E=10 and $j=1$; and finally


*A 15 with E=0,  and  $j=3$, $j=2$, and $j=1$.
I should agree with you that the traceless arrangement of the eigenvalues of B suggests a spin structure reminiscent of $L_3$, but I see no indication of the Casimir invariants $j(j+1) $ reflected anywhere in your energy eigenvalues. $L_3$ generates a U(1), a circle, like electric charge: I see no trace of the other two generators of SO(3) you mention. You can find operators in each block not commuting with $L_3$, which then connect the differing j multiplets, like the Laplace-Runge-Lenz generators in Hydrogen. But in the Hydrogen atom, the energy eigenvalue is inversely proportional to the degeneracy, $n^2=\sum_{j=0}^{n-1}(2j+1)$.
Unlike the SO(4) ~ SU(2) × SU(2) of the Hydrogen atom, however, I see no association of the energy eigenvalues with any Casimir invariant (what I was hoping for, when I asked you my comment question). Here, the energy is not linked to the degeneracy.
In the absence of more information, I see a different symmetry for each block: the 8 has an SO(8) symmetry, the 3 an SO(3), and the 15 an SO(15). The B sub-block in each is just one of the several generators of each group, not commuting with the rest. For instance, in the 3 block, the diagonal $L_3$ you see does not commute with the other two generators $L_1$ and $L_2$, etc.

Soft, touchy-feely note linked to your subsequent question/plea.
If you really wanted some type of a fantasy visualization of your 27 analogous to the infinite spectrum of hydrogen atoms, imagine some type of finite potential well with some type of spherical type symmetry. Your wave function, then, could be some factorized product of the type
$$
\psi(x)= L^j_E (r)   Y^m_j(\theta,\phi),
$$
where E is analogous to the principal quantum number n, except it now ranges over just 4 values, -2,0,4, and 10.
Like the Laguerre polynomials of hydrogen, its radial wavefunction depends not only on E, but on a "centrifugal barrier" j as well, where j can now take integer values circumscribed in the  mysterious way detailed above by E.  Each value j, in turn, represents a 2j+1 -dimensional multiplet, as you observed, and seen above. Finally, the larger over-all degeneracy of every energy level E is put together by varying js, organized by an over-arching mystical symmetry structure B, analogous to the SO(4) of Hydrogen.  Its 8,15,1,3 patter is for you to understand. I have no insights there...
