The above screenshot is from Takesaki's book(Vol 2, Chapter VIII) I wonder how to prove the implication (ii)→(i)?
By Theorem 2.11, we have $\sigma_t^{\psi}(x)=h^{-it}\sigma_t^{\varphi}(x)h^{it}$ for all $x\in M$. We have $\psi\circ\sigma_t^{\psi}(x)=\psi\circ\sigma_t^{\varphi}(h^{it}xh^{-it})$.
How to show that $\psi\circ\sigma_t^{\varphi}(x)=\psi(x)$?
The modular group is defined via the GNS representation for $\psi$. So \begin{align} \psi(\sigma^\psi_t(x))&=\langle \sigma_t^\psi(x)\Omega,\Omega\rangle =\langle \Delta^{-it} x\Delta^{it}\Omega,\Omega\rangle =\langle x\,\Delta^{it}\Omega,\Delta^{it}\Omega\rangle=\langle x\,\Omega,\Omega\rangle=\psi(x). \end{align} The equality $\Delta^{it}\Omega=\Omega$ follows from $\Delta\Omega=\Omega$, and this in turn follows by definition: $$ \Delta\Omega=FS\,\Omega=F\Omega=\Omega. $$