Well, it depends what are you looking on. What I've written down is the wave function, representing the state of a system. It's in fact independent of the picture used to calculate it. It's easy to see in the bra-ket formalism:
Heisenberg picture
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The state vector is time-independent:
$$|\psi(t) \rangle_{\text{H}}=|\psi(t_0) \rangle_{\text{H}},$$
while the eigenstates (e.g., the position eigenstates) are time dependent) according to
$$|x,t \rangle_{\text{H}}=\exp[\mathrm{i} \hat{H} (t-t_0)] |x,t_0 \rangle_{\text{H}}.$$
Thus the wave function (which is picture independent!) is given by
$$\psi(t,x)=_{\text{H}} \langle x,t|\psi,t \rangle_{\text{H}}=_{\text{H}} \langle x,t_0| \exp[-\mathrm{i} \hat{H} (t-t_0)|\psi,t_0 \rangle_{\text{H}}.$$
If the state vector is an energy eigenstate, you have
$$u_E(t,x)=\exp[-\mathrm{i}(t-t_0) t] u_E(t_0,x).$$
Schrödinger picture
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The state ket carries the full time dependence
$$|\psi,t \rangle_{\text{S}}=\exp[-\mathrm{i}(t-t_0) \hat{H}]|\psi,t_0 \rangle_{\text{S}},$$
and the eigenkets of the observable operators are time-independent
$$|x,t \rangle_{\text{S}}=|x,t_0 \rangle_{\text{S}}.$$
Again you find the same time dependence of the wave function (which is picture independent)
$$\psi(t,x)=_{\text{S}} \langle x,t|\psi,t \rangle_{\text{S}}=_{\text{S}} \langle x,t_0|\exp[-\mathrm{i}(t-t') \hat{H}] \psi,t_0 \rangle_{\text{S}}.$$
If the particle is in an energy eigenstate at ##t_0##, you find again the same result as when using the Heisenberg picture, as it must be
$$u_E(t,x)=\exp[-\mathrm{i} E (t-t_0)] u_{E}(t_0,x).$$
Of course, this tells us that the state doesn't change in this case, because the time evolution boils down to a simple phase factor, and the physical meaning is just the ##|\psi(t,x)|^2##, which is the probability distribution for finding the particle at ##x## when looking for it at time ##t##. For the energy eigenstate as the state vector, the phase factor cancels, and thus the energy eigenstates are precisely the stationary states of the particle.
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