A comment to this answer to another question states
I would imagine that for any linear non-unitary time-evolution operator, I can find a unitary one that will yield the same expectation values for every [physical state], which makes non-unitary time-evolution with manual normalization equal to unitary time evolution with standard normalization.
Is this correct?
No. For any particular initial state $|\psi_0\rangle$, we can manually normalize the hypothetical non-unitary but linear time-evolution operator $\hat{O}(t)$ in such a way that the manually normalized operator $\hat{O}_n(t) \equiv N_{\psi_0}(t)\, \hat{O}(t)$ produces a time-evolved trajectory $|\psi(t)\rangle = \hat{O}_n(t) |\psi_0\rangle$ with constant norm. But the key point is that the manual normalization function $N_{\psi_0}(t)$ necessarily depends on the particular initial state $|\psi_0\rangle$; there is not in general any manually normalized version of $\hat{O}(t)$ that preserves the norm along the trajectories for all initial states, like a unitary time-evolution operator does. The unitarity of the time-evolution operator is therefore a much stronger requirement than mere linearity, and you can't manually normalize an arbitrary linear time-evolution operator to a unitary one. (But note that the physical interpretation of unitarity is somewhat obscure in the projective-space formalism, where physical states don't have norms.)
As a simple example, consider the hypothetical linear but non-unitary time-evolution operator
$$\hat{O}(t) = \left( \begin{array}{cc} 1 & i \omega t \\ 0 & 1 \end{array}\right).$$
This operator trajectory is a one-parameter Lie group, i.e. it satisfies the composition property $\hat{O}(t_2) \hat{O}(t_1) = \hat{O}(t_2 + t_1)$. It preserves the norm of the initial state $(1, 0)$, so the manual normalization function for that initial state is the trivial $N(t) \equiv 1$. But the operator scales the norm of the initial state $(0, 1)$ over time as $\sqrt{1 + (\omega t)^2}$, so the manual normalization function for that initial state is $N(t) = 1/\sqrt{1 + (\omega t)^2}$. You can't normalize $\hat{O}(t)$ to simultaneously preserve the norm of both initial states. (Relatedly, the generator of this Lie group $$i \frac{d\hat{O}}{dt}|_{t=0} = \left( \begin{array}{cc} 0 & -1 \\ 0 & 0 \end{array}\right)$$ is not Hermitian.)
Given a linear (but possibly non-unitary) time-evolution operator $\hat O(t)$, "manual normalization" would mean to consider the time evolution $$ |\psi(t)\rangle = \frac{\sqrt{\langle \psi_0 | \psi_0 \rangle}}{\sqrt{\langle \psi_0 | \hat O(t)^\dagger \hat O(t) |\psi_0\rangle}} \, \hat O(t) |\psi_0\rangle . $$ It is clear that this map $|\psi_0\rangle \mapsto |\psi(t)\rangle$ is non-linear in general (except if $\hat O(t)^\dagger \hat O(t)$ is a multiple of the identity). In other words, we can only "repair" normalization at the cost of linearity.
