Suppose that $A$ is a separable, simple, non-unital C\*-algebra. Let $\varphi$ be an approximately inner automorphism on $A\otimes\cal K$, meaning that there exists a sequence of unitaries $v\_n$ in the multiplier algebra $\mathcal{M}(A\otimes\mathcal{K})$ with $v\_n x v\_n^\* \stackrel{n\to\infty}{\longrightarrow} \varphi(x)$ for all $x\in A\otimes\cal K$. We also denote by $\varphi$ the induced automorphism on the multiplier algebra.


Let $e\_{11}\in\cal K$ denote the obvious rank one projection.



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> **My question:** Is $1\otimes e\_{11}$ Murray-von-Neumann equivalent to $\varphi(1\otimes e\_{11})$ in the multiplier algebra $\mathcal{M}(A\otimes\mathcal{K})$?
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It is obvious that this is true if $A$ is unital, but unclear (to me) otherwise. I don't know whether simplicity of $A$ is a red herring here, but that is the case that I am interested in for now.



Induced map of an approximately inner automorphism on the multiplier algebra of $A\otimes\mathcal{K}$
If $A$ has stable rank one then this can be deduced from the following result:
If $B$ has stable rank one and $a,b\in B\_+$ are Cuntz equivalent then
the right ideals $\overline{aB}$ and $\overline{bB}$ are isomorphic as Hilbert modules. This is Theorem 3 of


Coward, Kristofer T.; Elliott, George A.; Ivanescu, Cristian. The Cuntz semigroup as an invariant for $C^\*$-algebras. J. Reine Angew. Math. 623 (2008), 161--193.


See also Proposition 1 of 


Ciuperca, Alin; Elliott, George A.; Santiago, Luis. On inductive limits of type-I $C^\*$-algebras with one-dimensional spectrum. Int. Math. Res. Not. IMRN 2011, no. 11, 2577--2615.


It's a little awkward to translate this into an answer to the question. Let me try.


First, this: Let $p,q\in M(B)$ be multiplier projections such that $pB\cong qB$ as Hilbert modules. Then $p$ is Murray-von Neumann equivalent to $q$. Proof: Let $v\colon pB\to qB$ be an isomorphism. Extend $v$ to $B$ setting it to 0 on $(1-p)B$. Then $v\in M(B)$ (regarded as the adjointable operators on $B$) and $v^\*v=p$, $vv^\*=q$.


Now let $B$ be separable and of stable rank one. Let $\phi$ be an approximately inner automorphism extended to $M(B)$. Let us show that any projection $p\in M(B)$ is Murray-von Neumann equivalent to $\phi(p)$. It suffices to show that $pB\cong \phi(p)B$. Let $c\in B$ be a strictly
positive element of $pBp$ (exists since $B$ is separable). Then $pB=\overline{cB}$.
Since $c^{1/n}\uparrow p$ strictly and $\phi$ is strictly continuous, we also have $\phi(p)B=\overline{\phi(c)B}$. But $c$ and $\phi(c)$ are approximately unitarily equivalent, so they are Cuntz equivalent. We can use the theorem recalled above.


To answer the question, we take $B=A\otimes \mathcal K$ and $p=1\otimes e\_{11}$.
Notice that to answer the question we only need $A$ $\sigma$-unital rather than separable.


Remark: There exist examples of positive elements $a,b$ in an $A$ (of sr=2) that are approximately unitarily equivalent and yet $\overline{aA}$ and $\overline{bA}$ are not isomorphic as Hilbert modules. However, (1) $A$ is non-simple and they way the examples work they can't be tweaked to get this, (2) The unitary conjugations moving $a$ to $b$ don't seem to converge pointwise to an automorphism. So there is a chance that the question has a positive answer without sr1.


