Can one operator commute with four other operators?

If an operator $A$ commutes with $X$ and $Y$ it already must be trivial ($A = c I, cin mathbb{C}$). One way to prove this is to note that $A$ also commutes with $Z$ since $Z = frac{1}{2}i(YX - XY)$, and $A$ trivially commutes with $I$. Since $A$ commutes with $I, X, Y, Z$, which is Pauli basis of all operators, then $A$ commutes with any other operator, hence it must be trivial.

In general case you can have any number of commuting operators since all diagonal operators commute, as was pointed out by Norbert Schuch.

In general, the answer is that you need to find the irreducible representation (irrep) structure of the algebra generated by ${M_1,M_2,...}$. This will tell you what the commutant of the algebra looks like (commutant = {all operators that commute with all operators of the algebra}) . Then you can pick $A$ from the commutant.

In many cases the irrep of the algebra is trivial, like when the generators act irreducibly on the whole space (example: ${X,Y}$ acting on the space of a qubit) . Then the commutan is just the operators proportional to the identity. Another simple case is when all ${M_1,M_2,...}$ commute with each other. Then you just need to simultaneously diagonalize them and the commutant is all operators that act arbitrarily on the degenerate sectors.

For the general case the irrep structure is hard to find. There are numerical algorithms out there that can do the job.

I am working on an algorithm that can find the irrep structure without numerics (if the irreps are not too complicated that is, but one can also run it numerically).