Excuse the long post, I just wanted to make the post self-contained so that those who don't have Murphy's book can also understand my question. Consider the following fragment in Murphy's book (p161):
Why is there a unitary $u$ in the unitisation $\tilde{A}$ with $\varphi(u)x = x_\rho$? Since we have an irreducible representation $(H_\rho, \varphi)$ of $\tilde{A}$, we know that $\varphi(\tilde{A})$ acts irreducibly on $H_\rho$. Since $x$ and $x_\rho$ are unit vectors, there is a unitary in $B(H_\rho)$ that takes $x$ to $x_\rho$, so by Kadison's transitivity theorem there is a unitary $\varphi(u) \in \varphi(\tilde{A})$ with $\varphi(u)x = x_\rho$. However, the book claims that we can choose $u$ to be a unitary in $A$. How can we ensure this? Is maybe the following true?
Let $(H, \varphi)$ be an irreducible representation of a unital $C^*$-algebra $A$. If $\varphi(a)$ is a unitary in $\varphi(A)$, then $a$ is a unitary in $A$?
I don't think this holds, but we do know $uu^*-u^*u \in \ker(\varphi)$
As mentioned in the OP, using Kadison's transitivity Theorem, one can find a unitary operator $v$ (called $\varphi(u)$ in the OP) in $\varphi(\tilde{A})$ with $vx = x_\rho$. By the very last sentence of KTT one may actually suppose that $v=e^{iw}$, for some self-adjoint operator $w$ in $\varphi(\tilde{A})$.
Next pick $z$ in $\tilde A$ such that $\varphi(z)=w$ and immediately replace it by $(z+z^*)/2$, so that the new $z$ is self-adjoint and still satisfies $\varphi(z)=w$. Finally set $u=e^{iz}$.
One then has that $u$ is unitary and $$ \varphi(u) = \varphi(e^{iz}) = e^{i\varphi(z)} = e^{iw} = v. $$