TITLE: Induced map on $K_{1}$-group
QUESTION [6 upvotes]: Let $A$ be a unital $C^{\ast}$-algebra. Any automorphism $\alpha$ of $A$ induces a map on $K_{1}(A)$ by $\alpha_{\ast}[u]=[\alpha(u)]$. 
Let the automorphism $\alpha$ be inner, does it follow that $\alpha_{\ast}$ is the identity map? 
I guess the answer is negative because unitary equivalence is far from homotopy of unitaries. The answer may also depend on the type of the $C^{\ast}$-algebra. Any suggestions?

REPLY [2 votes]: If I may add a little to the above answer, $K_1(A)$ is by definition the group of homotopy classes of unitaries in the unitization  $(K \otimes A)^{\sim}$ of its stabilization $K \otimes A$ (replace $A$ by $A^{\sim}$ in the non-unital case as julien stated above.) 
Polar decomposition shows that every unitary in $(K \otimes A)^{\sim}$ is homotopic to one of the form $u \oplus  (1 - 1_{M_n(A)} ) $ where $u \in M_n(A)$. 
Also, for any unitary $u$ in a corner $p (K \otimes A)^{\sim} p$ and for any partial isometry $s$ with $s^*s = p$, $u \oplus (1-p)$ is homotopic to $sus^* \oplus (1-ss^*)$. 
So every $K_1$ class can be represented by some unitary $u \in M_n(A)$ after adding $1$'s down the diagonal and you can move a unitary $u \in M_n(A)$ up and down the diagonal without changing its $K_1$-class. It is immediate that $K_1$ is abelian and inner-automorphisms descend to the identity map on $K_1$.