Let $(M,L^2(M),J, L^2(M)^{+})$ be the standard form of a von Neumann algebra $M$. $L^2(M)$ is endowed with the $M$-$M$ bimodule:$x\xi y=xJy^{*}J\xi$ for all $x,y\in M$ and $\xi \in L^2(M)$.
Proposition 4.1 Let $M$ be any factor with separable product, $\theta \in \operatorname{Aut}(M)$ any automorphism and $\omega \in \beta(\mathbf N) \setminus \mathbf N$ any nonprincipal ultrafilter. Consider the following properties:
(i) $\theta \in \overline{\operatorname{Inn}}(M)$.
(ii) There exists a unitary $u \in M^\omega$ such that $\theta(x) = u x u^*$ for all $x \in M$ and $\theta(\varphi)^\omega = u\varphi^\omega u^*$ for some (or any) faithful normal state $\varphi \in M_*$.
(iii) There exists a nonzero partial isometry $v \in M^\omega$ such that $\theta(x)v = v x$ for all $x \in M$ and $v\varphi^\omega = \theta(\varphi)^\omega v$ for some (or any) faithful normal state $\varphi \in M_*$.
(iv) There exists a nonzero partial isometry $v \in M^\omega$ such that $\theta(x)v = v x$ for all $x \in M$.
(v) The automorphism $\theta \odot \operatorname{id}$ of $M \odot M^\text{op}$ extends to an automorphism of the C*-algebra $\operatorname C^*_{\lambda\cdot\rho}(M)$ generated by the standard representation $\lambda\cdot\rho$ of $M \odot M^\text{op}$ on $\operatorname L^2(M)$.
Then we have (i) ⇔ (ii) ⇔ (iii) ⇒ (iv) ⇒ (v).
Proof We only prove (iv) ⇒ (v) since the other implications are well-known (for the implication (iii) ⇒ (ii), see the end of the proof of [9, Theorem 1] starting from Lemma 4). Let $v \in M^\omega$ be a nonzero partial isometry such that $\theta(x)v = v x$ for every $x \in M$. Note that $v v^* \in M' \cap M^\omega$ and so $\operatorname E_M(v v^*) = \lambda \in \mathbf R_+^*$. Take $T = \sum_i x_i \otimes y_i^\text{op} \in M \odot M^\text{op}$. For every unit vector $\xi \in \operatorname L^2(M)$, we have in $\operatorname L^2(M^\omega)$ the following equalities:
$$\left\lVert\sum_i \theta(x_i)\xi y_i\right\rVert \mathrel{\overset*=} \frac1\lambda\left\lVert v^*\sum_i \theta(x_i)\xi^\omega y_i\right\rVert = \frac1\lambda\left\lVert\sum_i x_i v^*\xi^\omega y_i\right\rVert.$$
Since
$$\lVert T\rVert_{\operatorname C^*_{\lambda\cdot\rho}(M)} \mathrel{\overset*=} \lVert T\rVert_{\operatorname L^2(M)} = \lVert T^\omega\rVert_{\operatorname L^2(M^\omega)}$$
we obtain
$$\left\lVert\sum_i x_i v^*\xi^\omega y_i\right\rVert \le \lVert v^*\xi^\omega\rVert\cdot\lVert T\rVert_{\operatorname C^*_{\lambda\cdot\rho}(M)} = \lambda\lVert T\rVert_{\operatorname C^*_{\lambda\cdot\rho}(M)}.$$
Thus we have shown that
$$\left\lVert\sum_i \theta(x_i)\xi y_i\right\rVert \le \lVert T\rVert_{\operatorname C^*_{\lambda\cdot\rho}(M)}$$
The Proposition is from the paper "Structure of bicentralizer algebras and inclusions of type III", which is a joint work of Ando, Haagerup, Houdayer and Marrakchi (doi:10.1007/s00208-019-01939-9, arXiv:1804.05706).
How to derive the two equalities which are marked ($*$). Is the first equality relevant to the faithful normal conditional expectation $E_M$?
If $x\xi y=\lambda (x)\rho(y)\xi, \forall x,y\in L^2(M)$. I cannot prove that the equality holds.
I was confused about the above equality for several days. I'll appreciate it if anyone give me some hint.