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Let $M_{\mathbb{Q}}$ be the universal UHF-algebra and let $\mathcal{O}_{\infty}$ be the infinite Cuntz algebra. Let $A$ be a Kirchberg algebra that satisfies the UCT with $K_0(A) \cong \mathbb{Q}^n$ and $K_1(A) = 0$. By the Kirchberg-Phillips classification theorem the projections $p_i \colon \mathbb{Q}^n \to \mathbb{Q}$ induce homomorphisms $A \otimes \mathbb{K} \to M_{\mathbb{Q}} \otimes \mathcal{O}_{\infty} \otimes \mathbb{K}$, which combine to give a homomorphism $$ \psi \colon A \otimes \mathbb{K} \to (M_{\mathbb{Q}} \otimes \mathcal{O}_{\infty} \otimes \mathbb{K})^n $$ that is the identity on $K$-Theory. Since $A \otimes \mathbb{K}$ is simple, $\psi$ is injective.

Given an automorphism $\alpha \in Aut(A \otimes \mathbb{K})$, is there an extension of $\alpha$ to an endomorphism $\beta_{\alpha} \in End((M_{\mathbb{Q}} \otimes \mathcal{O}_{\infty} \otimes \mathbb{K})^n)$ with $\psi \circ \alpha = \beta_{\alpha} \circ \psi$? If yes, can $\beta_{\alpha}$ be chosen such that $\beta_{\alpha \circ \alpha'} = \beta_{\alpha} \circ \beta_{\alpha'}$?

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This is definitely not an answer but too long for a comment.

The algebra on the right-hand side has a fairly simple primitive ideal space, and it may not be too hard to try to do what you want by hand, using Kirchberg's results on non-simple purely infinite classification. I'm not sure but you might be able to write the right-hand side algebra as some nice crossed product (using non-simple AF-algebras or non-pointwise outer automorphisms) that can help with this. Maybe a careful application of Kirchberg-Phillips will also work.

Whether you can choose $\beta$ to behave well under composition may turn out to be hard to prove. You are essentially asking whether actions on $K$-theory lift to actions on the algebra. Even on Kirchberg UCT algebras, this is tricky. In this paper http://arxiv.org/pdf/math/0608093.pdf Katsura does some things for some finite groups.

Hope this helps!

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    $\begingroup$ You surely know this, but Kirchberg's paper where he classifies non-simple purely inifinite algebras is in German. This shouldn't be a problem for you, though ;) $\endgroup$ Feb 7 '14 at 21:17

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