How does the probability of a system being in a state change with time in the Heisenberg picture if the state vector is time independent?
Any physical result must be independent of picture chosen. So the probability of measuring the system in state $i$ at time $t$ must be the same in, e.g., the Heisenberg picture as in the Schroedinger picture. The funky thing about the Heisenberg picture is that the eigenkets evolve with time, unlike in the Schroedinger picture.
In equations, the probability of measuring a system represented by $|\psi(t)\rangle$ in state $|i\rangle$ in the Schroedinger picture is related to their overlap, $\langle i|\psi(t)\rangle = \langle i|U(t,t_0)|\psi(t_0)\rangle$, where $U$ is the time evolution operator.
In the Heisenberg picture, the state of the system is a constant in time. But because operators have time dependence, their eigenstates pick up a time dependence. In particular, the probability of measuring a system represented by $|\psi\rangle$ in state $|i(t)\rangle$ in the Heisenberg picture is related to their overlap, $\langle i(t)|\psi\rangle = \langle i(t_0)|U(t,t_0)|\psi(t_0)\rangle$, where $U$ is again the time evolution operator.
One can see then that the two results agree, as they must.