 Solution

c1 is true for density-state controllability. Every qubit density operator can be written $\rho ={\lambda }_{2}I+({\lambda }_1-{\lambda }_2)|\psi ⟩⟨\psi |$ using its eigenvalues. A control taking one pure orthogonal projector to any other therefore also connects every pair of qubit density operators with the same spectrum. In Lie-algebra language, $\mathfrak{sp}(1)=\mathfrak{su}(2)$; there is no larger-dimensional symplectic exception. This does not supply arbitrary global phase: $SU(2)$ alone is already a counterexample to an implication involving strict operator controllability.

c2 is false. For every even $N=2\ell \ge 4$, the defining $Sp(\ell )$ action is transitive on unit vectors, so a system generating this algebra is pure-state controllable. It still preserves additional mixed-state invariants and lacks density operator controllability, as the preceding explicit four-dimensional example demonstrates. pure states have fewer orbit requirements than general mixed states.