Theorem 2.11 is from Takeski's book (Vol 2. Chapter VIII, Section 2).
It mentions that we can get $\sigma^{\psi}_t(x)=(he_n)^{it}\sigma^{\varphi}_t(x)(he_n)^{-it}$ by Lemma 2.10. I have a question: $x\in M_{e_n}$, it is also in $M$, by Lemma 2.10, we have $\sigma^{\psi}_t(x)=(he_n)^{it}\sigma^{\varphi}_t(x)(he_n)^{-it}$$\sigma^{\psi}_t(x)=(h)^{it}\sigma^{\varphi}_t(x)(h)^{-it}$. Why should we replace $h$ by $he_n$.
Another question: how to derive $(he_n)^{it}=(h)^{it}$?
In the proof of Theorem of
The fact that $h$ is affiliated with $\mathcal M_\varphi$ implies that $he_n\in \mathcal M_\varphi$. So, for $x\in e_n\mathcal M e_n$ \begin{align} \sigma_t^\psi(x) &= (he_n)^{it}\sigma_t^\varphi(x)(he_n)^{it}=h^{it}e_n\sigma_t^\varphi(x)e_nh^{it}\\[0,3cm] &=h^{it}\sigma_t^\varphi(e_n)\sigma_t^\varphi(x)\varphi(e_n)h^{it}=h^{it}\sigma_t^\varphi(e_nxe_n)h^{it}\\[0.3cm] &=h^{it}\sigma_t^\varphi(x)h^{it}. \end{align} The first equality is due to Lemma 2.10 applied to $he_n\in e_n\mathcal Me_n$.