The time translation is given by a finite unitary transformation $$\hat{U_{\tau}}(\hat{H}) = e^{\big(\frac{i}{\hbar}\tau \bar{H}\big)}.$$ Where $$\hat{U_{\tau}}(\hat{H})|\psi(t) \rangle = |\psi(t-\tau)\rangle).$$ We can show that the hamiltonian is invariant under the finite unitary translation given by $$\hat{U}(t,t_0) = e^{-\frac{i(t-t_0)\hat{H}}{\hbar}}$$ so we have $\hat{H} = e^{\frac{-i(t-t_0)\hat{H}}{\hbar}} \hat{H} e^{\frac{i(t-t_0)\hat{H}}{\hbar}} = \hat{H}$. Aslo we have that $[\hat{H}, \hat{H}] = 0$, thus $$\frac{d}{dt} \langle \hat{H} \rangle = \frac{1}{i \hbar} \langle [ \hat{H}, \hat{H} ] \rangle + \langle \frac{\partial \hat{H}}{\partial t} \rangle = 0.$$
So $\hat{H}$ is conserved.
How does all of this imply that if $\psi(t)$ satisfies the time dependent Schrodinger equation then so does the time displaced $\psi(t- \tau)?$