Pure isometries are unitarily equivalent to a shift (Wold decomposition). Do the corresponding intertwining relations also hold for the adjoints? Let $T \in L(H)$ be an isometry on a Hilbert space. Further assume that $T$ is pure, i.e. $T^{*m} \xrightarrow {m \rightarrow \infty} 0$ in the strong operator topology, or equivalently, there is no unitary part in the Wold decomposition of $T$. In other words, by the Wold decomposition Theorem, there is a Hilbert space $D$ and a unitary $V: H \rightarrow H^2(\mathbb{D}, D)$ such that $VT = M_z V$ holds. (Here, $H^2(\mathbb{D}, D)$ denotes the $D$-valued Hardy space on $\mathbb{D}$ and $$M_z: H^2(\mathbb{D}, D) \rightarrow H^2(\mathbb{D}, D), (M_zf)(z) = z f(z)$$ the shift operator on it) Is it true that we also have $VT^* = M_z^* V$ in this case?

Yes, because $V$ is a unitary. This allows you to write the intertwinning relation as $$ VTV^*=M_z. $$ Now you can take adjoints and multiply by $V$ on the right.