Is the center of the automorphism group of a von Neumann algebra M trivial whenever M is a factor? 
Question: Is the center of the automorphism group of a von Neumann algebra $\mathscr{M}$ trivial (=$\{\mathrm{id}\}$) whenever $\mathscr{M}$ is a factor (=$\mathscr{M}$ has center $\{\lambda I; \lambda \in \mathbb{C}\}$)?

It is true in the finite dimensional case. A finite dimensional factor is $M_n(\mathbb{C})$ for some $n \in \mathbb{N}$.
We proceed in two steps.  First, every automorphism of $M_n$ is of the form $a \mapsto u^*a u$ for some unitary $u$.  We will show that if an automorphism is in the center of $\mathop{\mathrm{Aut}}$, then $u$ is in the center of its unitary group.  Finally, we show that the center of the unitary group is "trivial".


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*There is a surjective group homomorpism $\varphi\colon U(n) \to \mathop{\mathrm{Aut}}(M_n)$ with $\varphi(u)(a)=u^*au$.
The kernel of $\varphi$ are the unitaries in the center of $M_n$. Thus $\ker \varphi = \{\lambda I;\ |\lambda|=1\}$.  Now, given any $u$ such that $\varphi(u)$ is in the center of $\mathop{\mathrm{Aut}}(M_n)$. Then for any other $v$ we have $vuv^{-1}u^{-1} \in \ker\varphi$. Thus $vuv^{-1}u^{-1}=\lambda I$ for some $\lambda \in \mathbb{C}$ with $|\lambda|=1$. In particular, when $u=v$ we find $I=\lambda I$. Apparently $vuv^{-1}u^{-1}=I$. Thus $u$ is in the center of $U(n)$.

*The embedding $e\colon U(n) \to GL(n)$ is an irreducible group representation.  A unitary $u$ in the center of $U(n)$, is an intertwining map from this representation to itself. Thus, by Schur's lemma, it is $\lambda I$ for some $\lambda \in \mathbb{C}$.  Clearly $|\lambda| = 1$. Thus the center of $U(n)$ is the kernel of $\varphi$.
 A: Suppose that $\phi $ is in the centre of $\text {Aut}(M) $. Fix a unitary $u\in\mathcal M $. Then $$\tag {1}\phi (uxu^*)=u\phi(x)u^*$$  for all $x $. In particular, when $x=u $, we have $\phi ( u)=u\phi (u)u^*,$ or $\phi (u)u=u\phi (u)$. As $u $ is unitary, this also implies that $u^*\phi (u) =\phi (u)u^*$.
Replacing $x $ with $xu $ in $(1) $, we get $\phi (ux)=u\phi (xu)u^*$, so 
$$u^*\phi (u)\phi (x)=\phi (x)u^*\phi (u). $$ As $\phi $ is onto, we deduce that $u^*\phi (u) $ is in the centre of $\mathcal M $.
Since $\mathcal M $ is a factor, $\phi (u)=\lambda_u  u $ for some $\lambda_u \in\mathbb C $. It is easy to see that the map $u\longmapsto\lambda_u $ is a group homomorphism.
(Thanks to Jesse Peterson for the following argument) From $\phi(u)=\lambda_u u$, we get that $\sigma(u)=\lambda_u\,\sigma(u)$, since $\phi$ preserves the spectrum. So, if the spectrum of $u$ has no rotational symmetry, then $\lambda_u=1$. The set of unitaries with no rotational symmetry is dense in the set of unitaries of $\mathcal M$ (via the Spectral Theorem, since we can use unitaries with finite spectrum and tweak the eigenvalues slightly so that there is no rotational symmetry). By continuity, it turns out that $\phi$ is the identity on all the unitaries, and so $\phi$ is the identity. 
