Let $U$ be the bilateral shift operator in $l^2(Z)$, and let $V$ stand for a rotation on an irrational angle $\alpha$ in $L^2(T, \mu)$, where $T$ is a circle with a rotation-invariant Lebesgue measure $\mu$. The spectra of both operators coincide as sets, but the spectrum of $U$ is absolutely continuous, while that of $V$ is pure point. Hence, no unitary equivalence between two operators is possible.

However, intuitively $U$ and $V$ are related in the followng way. Let's extend the action of $V$ to the space of tempered distributions ${\cal S}'(T)$. Then, $V$ acts on the set of vectors $e_n=V^n \delta$ exactly by a bilateral shift (here $\delta$ stands for the Dirac delta at some point of $T$). Unfortunately, there is no obvious way to identify the closure of $\mathrm {span} (e_n)$ in ${\cal S}'(T)$ with $l^2(Z)$. Thus the question is: is it possible make this naive argument rigorous, say by extending the action of $U$ and $V$?