Say I have a Lie algebra L. I then construct it's universal enveloping lie algebra U(L). Let $F^i(U(L))$ form a filtration i.e for $i < j$ from some indexing set I, $F^i(U(L)) \subset F^j(U(L))$ and $F^i(U(L))\cdot F^j(U(L)) \subset F^{i+j}(U(L))$. Let's say I have a lie group G acting on U(L) as $g^{-1}xg$ where $g \in G $ and $x \in F^i(U(L))$ (I think I would like $g^{-1}xg$ to remain in $F^i(U(L))$). What group acts on the space $F^i(U(L))/F^{i-1}(U(L))$ and leaves it fixed. We could possibly relax this condition and say it leaves $F^i(U(L))$ fixed for any i.
This I think is a concrete example from physics: Say, I have a qubit Hamiltonian $H = \sum_{p,n}^N A^{pn} a^\dagger_p a_{n} + \sum_{jklm} B^{jklm}a^\dagger_{j}a^\dagger_{k}a_{l }a_{m} $ + h.c, where the operator obey a commutation relation. Now is it possible to find unitaries $U_r$ such that $U^{\dagger}_ra_jU_r =\sum S_j^i a_i $. So if $U_r$ acts like $U_r^{\dagger}HU_r$, the form of the hamiltonian is preserved we will just have different constant $A'^{pn}, B'^{jklm}$. It turns out if the operators are fermonic then the $U_r$ exist and $S \in SU(N)$.
It seems what needs to happen is that the $U_r$ should keep a two body interaction a two body interaction and a 4 body interaction a 4 body interaction. In the fermonic case the 2 body interaction can be written as a sum of a two body term and a constant (from anti-commutation relation ) and the 4 body can be written as 4 body term plus 2 two body term (from anti-commutation). Now the hamiltonian is an element in the enveloping algebra and it seems to me that $U_r$ must 'respect' the sets in filtration used by the Hamiltonian. I have tried to guess what 'respect' means i.e that the summands in graded algebra formed from the universal enveloping algebra are left fixed by $U_r$.