Let $\mathfrak{A} \subset B(H)$ be a C*-algebra. For $x \in H$, $\|x\|=1$, define a vector state $\omega_x$ on $\mathfrak{A}$ by $\mathfrak{A} \ni A \mapsto \langle x, Ax \rangle$. Assume $(x_n)_{n\in \mathbb{N}} \subset H$ is a sequence of cyclic vectors, and that $\omega_{x_n}$ converges in norm to $\omega_x$ for some cyclic vector $x\in H$.
In general we cannot conclude that $x_n \to x$ from the above convergence because different vectors from $H$ may generate the same vector state (e.g. $\omega_x = \omega_{Ux}$ for a unitary $U \in \mathfrak{A}'$ in the commutant of $\mathfrak{A}$). However, we also know from the uniqueness of the GNS representation that if $\omega_y = \omega_z$ for two normalised cyclic vectors $y, z \in H$, then a unitary operator $U\in\mathfrak{A}'$ exists such that $Uy=z$.
Question: Is it possible to fix a sequence of unitary operators $(U_n)_{n\in\mathbb{N}} \subset \mathfrak{A}'$ such that $U_n x_n \to x$?
I am able to answer the above question affirmatively in case the projection $P_x = \langle x, \cdot \rangle x$ is an element of $\mathfrak{A}$. For this, observe that $$|\langle x, x_n \rangle|^2 = \omega_{x_n}(P_x) \to \omega_{x}(P_x) = \| x\|^2 = 1.$$ This implies $\mathrm{e}^{\mathrm{i}\alpha_n} \langle x, x_n \rangle \to 1$ for suitably chosen phases $\mathrm{e}^{\mathrm{i}\alpha_n}$. Moreover, we may write $x_n$ as $$x_n = x_n^\perp + \langle x,x_n \rangle x,$$ where $x_n^\perp$ is orthogonal to $x$. It is not difficult to verify $x_n^\perp \to 0$. Thus, by Pythagoras, $$\| \mathrm{e}^{\mathrm{i}\alpha_n} x_n - x \|^2 = \| x_n^\perp \|^2 + \| (\mathrm{e}^{\mathrm{i}\alpha_n} \langle x, x_n \rangle - 1)x \|^2 \to 0.$$ However, this argument relies on $P_x \in \mathfrak{A}$. This is not true for general $C^*$-algebras.