How do you find the missing symmetry? I am considering a system that has degenerate energies, degenerate $S^{2}$, and degenerate $S_{z}$, so how do I find the missing symmetry, probably rotational, that makes the states unique?
For example,
$$
|1,1\rangle_1 = \frac{1}{2}\bigg[ |\uparrow \downarrow \uparrow \uparrow \rangle + |\downarrow \uparrow \uparrow \uparrow \rangle - |\uparrow \uparrow \uparrow \downarrow \rangle - |\uparrow \uparrow \downarrow \uparrow \rangle \bigg]
$$
and
$$
|1,1\rangle_2 = \frac{1}{\sqrt{2}}\bigg[ |\uparrow \uparrow \uparrow \downarrow \rangle - |\uparrow \uparrow \downarrow \uparrow \rangle \bigg]
$$
and
$$
|1,1\rangle_3 = \frac{1}{\sqrt{2}}\bigg[ |\uparrow \downarrow \uparrow \uparrow \rangle - |\downarrow \uparrow \uparrow \uparrow \rangle \bigg]
$$
I want to do matrix representation, so I have the $16\times16$ matrix, but I'm not sure how to populate it. I'm assuming I find eigenvalues?
 A: Assuming you are looking at all four $|n,1\rangle$ states,
and defining
$$
|2,1\rangle_0 = \frac{1}{2}\bigg[ |\uparrow \downarrow \uparrow \uparrow \rangle + |\downarrow \uparrow \uparrow \uparrow \rangle + |\uparrow \uparrow \uparrow \downarrow \rangle + |\uparrow \uparrow \downarrow \uparrow \rangle \bigg]
$$
as the only state that is not annihilated by a raising operator, so it is part of the quintet (spin 2), your four orthogonal states are differentiated by the projection operators where () means symmetrizing, and [] means antisymmetrizing, and the numbers denote original  spin locations.
So,
$$
P_{(1234)} |2,1\rangle_0 = |2,1\rangle_0 , ~~~0 ~~~\hbox {for the rest}, \\
P_{[1,3][2,4](12)(34)}|1,1\rangle_1= |1,1\rangle_1,  ~~~0 ~~~\hbox {for the rest}, \\
P_{(12)[34]} |1,1\rangle_2=  |1,1\rangle_2 , ~~~0 ~~~\hbox {for the rest}, \\ 
P_{[12](34)}|1,1\rangle_3= |1,1\rangle_3 , ~~~0 ~~~\hbox {for the rest}.
$$
I 'm sure my notation is unconventional and redundant, but the proper notation is in discussions of Young tableaux for the symmetric croup. I'm just stating in Pidgin the symmetries and anti symmetries that uniquely distinguish your four states. Check they are automatically orthogonal to each other. (The 0th one, of course, is distinguished by its $S^2$ eigenvalue, so, ultimately, it is not part of your problem.)
You may construct the 4x4 orthogonal matrix that  encodes that projection,  acting on the relevant vectors $( |\downarrow \uparrow \uparrow \uparrow \rangle , |\uparrow \downarrow \uparrow \uparrow \rangle ,  |\uparrow \uparrow \downarrow \uparrow \rangle ,|\uparrow \uparrow \uparrow \downarrow \rangle )^T$,
$$\begin {pmatrix} 1/2 & 1/2 &1/2 &1/2 \\
1/2 &1/2 &-1/2 &-1/2 \\
0&0&-1/\sqrt{2} & 1/\sqrt{2}\\
-1/\sqrt{2}& 1/\sqrt{2} &0&0
\end{pmatrix},$$
and manifestly displays the permutation symmetries in its rows.
A: This is too long for a comment so…
What you want to do is check if the states transform as a linear combinations of themselves under the permutation group $S_4$ of $4$ objects.  This group has $4!$ elements so it might be heavy going but since any element in $S_4$ is a product of transpositions (elements which permute only two particles and leave two alone), you are good enough to check for the transpositions $P_{12}, P_{13}, P_{14},P_{23},P_{24}, P_{34}$.
The hypothesis here is that your 3 basis states span an irreducible representation of $S_4$.  Indeed, there is one irrep (denoted by the partition $\{3,1\}$ of $S_4$) that is of dimension 3, so the odds are in your favour.
If indeed it is the case that your states do for a basis for the irrep $\{3,1\}$, you might then try to organize them so they span an irrep of the subgroup $S_3$, which would permute the first three entries only, and then from there continue with the subgroup $S_2$ which permutes the first two entries.  This would allow you to label your states by an irrep of $S_4$ (the irrep is probably $\{3,1\}$), simultaneously with an irrep of $S_3$ and simultaneously with an irrep of $S_2$.
If indeed your current set spans $\{3,1\}$ of $S_4$,  you can obtain the $S_3$ irreps in $\{3,1\}$ using the standard branching rules, or by using the Young diagram method.  The possible irreps of $S_3$ inside this irrep of $S_4$ are $\{3\}$ and $\{2,1\}$.  $\{3\}$ is the fully symmetric irrep of $S_3$ and of dimension $1$ while $\{2,1\}$ is of mixed symmetry and of dimension $2$.
The procedure is tedious.  Knowing that the irrep $\{3\}$ of $S_3$ will appear if what you have is $\{3,1\}$ of $S_4$, you might start by constructing a combo of the states that you have that will be fully symmetric under $S_3$, but it’s still tedious.
