Can the intersection of a unitary and an irreducibly represented injective $C^*$-algebra be $\{0\}$? Let $\mathcal{A}$ be an injective $C^*$-algebra irreducibly acting on a Hilbert space $\mathcal{H}$, and let $\phi$ be a completely positive idempotent from $\mathbb{B}(\mathcal{H})$ onto $\mathcal{A}$, and let $u\in\mathbb{B}(\mathcal{H})$ be a unitary.

Question: Can we conclude that $\phi(u)\ne0$? If necessary, we can assume that $u$ normalizes $\mathcal{A}$, that is, $u\mathcal{A}u^*=\mathcal{A}$.

The point is the irreducibility, since without it $\phi(u)$ can easily be $0$. The question appears to be easy, but I have been struggling for months, and I have already asked five experts.
 A: No. Take any irreducibly represented simple and injective C*-algebra $A\subset B(H)$ and an outer automorphism $\theta$ of period $2$ (which is easy to find when $A$ is the hyperfinite II_1 factor).
Case 1: The representation is covariant, i.e., $\exists u\in{\mathcal U}(H)$ such that $uau^*=\theta(a)$ for $a\in A$. We may assume $u^2=1$ by irreducibility. Since $\theta$ is outer, $A\rtimes_\theta({\mathbb Z}/2{\mathbb Z})$ is simple by Kishimoto's theorem. Hence, $\mathrm{C}^*(A,u)\cong A\rtimes_\theta({\mathbb Z}/2{\mathbb Z})$ and there is a conditional expectation $\phi\colon\mathrm{C}^*(A,u)\ni a+bu\mapsto a\in A$, $a,b\in A$. The map $\phi$ extends on $B(H)$.
Case 2: The representation is not covariant, in which case there is no nonzero $x\in B(H)$ such that $xa=\theta(a)x$ for all $a\in A$, by Schur's lemma.  Consider $A_1:=A\rtimes_\theta({\mathbb Z}/2{\mathbb Z})=\mathrm{C}^*(A,v)$ represented on $H\oplus H$, where $a\in A$ acts by $a\oplus\theta(a)$ and the switching unitary operator $v(\xi\oplus\eta)=\eta\oplus\xi$ implements $\theta$. Now, $A_1$ is injective, irreducibly represented on $H\oplus H$, and the dual automorphism $\hat{\theta}\colon a+bv\mapsto a-bv$ is outer and implemented by the unitary operator $u:=1\oplus-1$ on $H\oplus H$. Hence it reduces to Case 1.
