Here's an argument (assuming global choice) that the codomain fibration will always be split in "natural" cases. Perhaps it's well-known, or perhaps it has an important shortcoming (such as the use of choice?). Or perhaps there's an error -- strictification is tricky!

**Claim:** Let $C$, $D$ be categories with pullbacks, and suppose there is a functor $p: C \to D$ with the following properties:

$p$ preserves pullbacks.

$p$ is an isofibration, and induces an isofibration on all slices (perhaps this is redundant?).

$p$ reflects identities (i.e. if $f: c \to c$ is an endomorphism in $C$, and $pf: pc \to pc$ is an identity, then so is $f$).

There is a cardinal $\kappa$ such that for every $f: x \to c \in C^{[1]}$, the isomorphism class $[f]$ of $f$ in the fiber category $p^{-1}(pf) \subseteq C/c$ has cardinality $\kappa$.

The claim is that if $D$ admits strict reindexing, then so does $C$, and $p$ preserves the strict reindexing.

**Upshot:** Most "natural" categories $C$ admit such a functor $p: C \to \mathsf{Set}$, with $\kappa$ being the cardinality of the universe (perhaps with an exception for objects over $\emptyset$, which often doesn't matter because there are few maps into $\emptyset$). I'm assuming that $\mathsf{Set}$ admits strict reindexing (right?). So most "natural" categories do as well.

**EDIT:** $\mathsf{Set}$ does indeed admit strict reindexing. To see this, as in the comment below, note that the fibration $\mathsf{Set}^{(-)} \to \mathsf{Set}$ admits a splitting, and the fibration $\mathsf{Set}/(-) \to \mathsf{Set}$ is isomorphic as a fibered caegory -- the fact that it's equivalent as a fibered category is standard, and each isomorphism class in each fiber category has the cardinality of the universe, in both fibrations, so an isomorphism of fibrations exists. So $\mathsf{Set}/(-) \to \mathsf{Set}$ also admits a splitting as desired.

So indeed, most "natural" categories admit a strict reindexing. Of course, it's completely undecideable to compute it because the choice function must solve the isomorphism problem!

**Proof of Claim:** Let $()^\ast$ be a system of strict reindexing in $D$. Fix enumerations of the isomorphism classes of the fibers of $p$ by $\kappa$ (using condition (4) and choice). Now, if $\gamma : c' \to c$ is a morphism in $C$ and $f: x \to c \in C/c$, take an arbitrary pullback of $f$ along $\gamma$. Because $p$ is an isofibration (condition (2))and $p$ preserves pullbacks (condition (1)), we can correct this choice to lie over the canonical pullback $(p\gamma)^\ast(pf)$. Moreover, we can choose a representative which has the same index under our enumeration as $f$ does; let's call it $\gamma^\ast(f)$. It remains only to nail down the lift of the other leg of the pullback square, but any two choices differ by an automorphism of $\gamma^\ast(f)$ which lies over the identity on $(p\gamma)^\ast(pf)$, so by condition (3) this choice is uniquely determined.

It's now obvious that the lifted operation $()^\ast$ is strictly functorial, because it preserves the indexing by our enumerations.