Let $\mathcal{H}$ be a separable complex Hilbert space. Let $(u_x)_{x\in X},(v_x)_{x\in X}$ be two families of vectors in $\mathcal{H}$ such that for any $x,x'$, $$\langle u_x,u_{x'} \rangle=\langle v_x,v_{x'}\rangle.$$
Then, there exists a unitary map $U\in \mathcal{U}(\mathcal{H})$ such that for any $x\in X$, $U(u_x)=v_x$.
Now consider families of such families indexed by $y\in Y$, that is $(u_{x,y}),(v_{x,y})$ such that for all $x,x'\in X,y\in Y$, $$\langle u_{x,y},u_{x',y} \rangle=\langle v_{x,y},v_{x',y}\rangle.$$ Then for any $y\in Y$, there exists a unitary map $U_y$ such that $U_y(u_{x,y})=v_{x,y}$.
If furthermore, we assume $X,Y$ to be measurable spaces and $u,v:X\times Y\to \mathcal{H}$ measurable, is there a choice of $U_y$ such that $y\mapsto U_y$ is measurable as a map from $Y$ into $\mathcal{U}(\mathcal{H})$ endowed with the Borel $\sigma$-algebra of the strong opearator topology ? (given that $\mathcal{H}$ is separable this should be equivalent to $y\mapsto U_y\xi$ measurable for any $\xi\in \mathcal{H}$).
If not, perhaps one can put stronger conditions on $X,Y$ such as being standard Borel spaces ?