Suppose I have a vector field $u: \mathbb{R}\to \mathbb{C}^n$ whose inner products satisfy $$(u_x,u_y) = f(x-y)$$ i.e. the inner product only depends on the difference of positions. Does there necessarily exist a set of unitary operators $L: \mathbb{R} \to \mathcal{L}(\mathbb{C}^n)$ such that $u_{y+x} = L_x u_y$ for all $x,y$?
If this is true, we certainly have $(u_{x+a}, u_{y+a}) = (L_a u_{x},L_a u_y) = (u_{x},u_y)$ and hence the inner product only depends on the difference of positions. I haven't been able to prove the converse (I tried contradiction) or come up with a counterexample. I'm also interested in replacing $\mathbb{R}$ with a general metric space, and my motivation is from condensed matter physics where $u_x$ is a Bloch function on the Brillouin zone. Note: the operators would also satisfy $[L_x,L_y] = 0$ and $L_xL_y = L_{x+y}$, like translation operators.
Note: To avoid some trivial counterexamples I'll simply ask that $L_{x}^\dagger L_x = I|_U$ where $U = \text{Im}(u)$, so $L$ need not be invertible.