In the book "Quantum Field Theory of Point Particles and Strings" by Hatfield, in pages 114-118 the author discusses the proper relation between interacting fields and in/out fields and the spectral representation of the two-point function. In particular \begin{eqnarray} \lim_{t\to -\infty}\varphi(x)&=&\sqrt{Z}\varphi_{\rm IN}(x)\\ \lim_{t\to \infty}\varphi(x)&=&\sqrt{Z}\varphi_{\rm OUT}(x)\tag{7.10} \end{eqnarray} and then studying the spectral representation one finds that $iZ$ is the residue of the propagator at the pole corresponding to the mass of the particle. Moreover the author shows that $0 < Z \leq 1$ and that $Z=1$ if and only if $\varphi(x)=\varphi_{\rm IN}(x)$. In page 118 he further comments that $Z=1$ implies $\varphi=\varphi_{\rm IN}$ and we end up with a free field theory.
Now consider the renormalized Lagrangian for the renormalized field $\varphi^R(x)$. In the on-shell scheme one sets the residue of the propagator at the mass pole to be $i$. In particular, comparing to the above discussion, this means that for the renormalized field $\varphi^R(x)$ we have one $Z^R=1$.
But now this is rather confusing because Hatfield says that a field for which the residue of the propagator at the mass pole is $i$ coincides with in/out fields and is a free field. But the renormalized field certainly can't be free (we are in the interacting theory after all!).
So what is reallly going on here? How can the renormalized field $\phi^R(x)$ have $Z^R=1$ and this still be an interacting theory? How to reconcile this renormalization condition with the remarks by Hatfield that a field whose corresponding $Z=1$ is free?