Corollary 4.8 in Awodey's book states that every functor $\mathcal F$:$\mathcal C$$\rightarrow$$\mathcal D$ factors $\mathcal F$ = $\mathcal H$$\mathcal G$ where

$\mathcal G$ : $\mathcal C$$\rightarrow$$\mathcal C/ker(F)$ is bijective on objects quotient functor and $\mathcal H$ : $\mathcal C/ker(F)$$\rightarrow$$\mathcal D$ is fully faithful. That is the composite ;

$\mathcal F$:$\mathcal C$$\rightarrow$$\mathcal C/ker(F)$$\rightarrow$$\mathcal D$.

We can also factorize $\mathcal H$ in the way $\mathcal F$ was factorized to obtain $\mathcal H$=$\mathcal H_1$$\mathcal G_1$ initializing at $\mathcal H_0$= $\mathcal H$ and $\mathcal G_0$= $\mathcal G$.

Question: upon iterated factorization of $\mathcal H_n$, does this finally converge on $\mathcal D$. And if so, what factorization system generally captures this ? My guess would be reflective-factorizaton systems or monadic decompositon but am looking for a grounded answer.

EDIT(4/4/2021): There seems to be two contradictory answers as seen from the comments:

(i) In the factorization $\mathcal H$=$\mathcal H_1$$\mathcal G_1$, $\mathcal H_1$=$\mathcal H$ and $\mathcal G_1$ is the identity on $\mathcal C/ker(F)$

(ii) An orthogonal factorization system ($\mathcal L$, $\mathcal R$) where the class $\mathcal L$ is of iterated strict-localizations(= iterated quotients) and $\mathcal R$ is the class of conservative functors. It is the factorization of a functor where the iterated quotient converges at the colimit of the transfinite sequence. See relevant links in comments.

How does one unify these two perspectives into one? My current intuition on the problem is that the transfinite sequence is an isomorphism/equivalence(by cancellation laws) and every functor is an iterated strict-localization. Is this true or does it collapse to the first argument?

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    $\begingroup$ In the factorization $\mathcal{H}=\mathcal{H}_1\circ\mathcal{G}_1$, isn't $\mathcal{H}_1=\mathcal{H}$ and $\mathcal{G}_1$ the identity functor on $\mathcal{C}/\ker(F)$? $\endgroup$ Mar 25 at 11:57
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    $\begingroup$ Jeremy is right. If you factor a fully faithful functor in this way, then the bijective-on-objects first factor will also be fully faithful (since fully faithful functors cancel on one side), and hence an isomorphism. $\endgroup$ Mar 25 at 16:05
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    $\begingroup$ This is actually an orthogonal factorization system: ncatlab.org/nlab/show/%28bo%2C+ff%29+factorization+system, which entails that factorizations are unique up to isomorphism. $\endgroup$ Mar 25 at 16:06
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    $\begingroup$ I am not sure what you think this is a counterexample to. These are two different factorization systems on Cat. In one of them the factorization can be constructed by hand in one step. In the other, there is a naive one-step factorization that is not the OFS factorization but can be used to construct it by a transfinite iteration. $\endgroup$ Apr 4 at 15:27
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    $\begingroup$ In the Corollary 4.8 that you cite, I see only that the functor you call $\mathcal H$ is faithful, not necessarily fully faithful. $\endgroup$ Apr 4 at 15:56

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