TITLE: Does there exist a terminal surjective discrete fibration out of $C$?
QUESTION [6 upvotes]: Let $DF$ denote the category whose objects are categories and whose morphisms $F\colon R\to S$ are the discrete fibrations. This category has applications to the real-world problem of structuring data. You can think of any discrete fibration $R\to S$ as providing a schematic structure $S$ for more raw data $R$. This question is about finding a best such structure for given data.
For any category $R$, one can define the coslice category 
$$DF_{R/}$$ of discrete fibrations out of $R$. Given a discrete fibration $R\to S$, call $S$ the base space. I want to find a minimal base space $R^{min}$ for a given $R$. To get at that idea, I'll ask for some kind of universal object.
I once asked on the categories mailing list whether $DF_{R/}$ has a terminal object in general. As was explained to me by Mark Weber and Thorston Palm, it does not. Basically, if $R=\emptyset$ then we're asking for a terminal object in $DF$, and by cardinality arguments, this does not exist. 
But in fact I was asking the wrong question. If I want minimal models, I actually want my discrete fibrations to be surjective. The counter-examples provided to me by Weber and Palm fail to cause a problem in that context. So here's the question. 
Question: Define $DFS$ to be the subcategory of $DF$ in which the morphisms are discrete fibrations $F\colon R\to S$, required to be surjective on objects. Then does $DFS_{R/}$ have a terminal object $R^{min}$ for a given category $R$?

Edit provenance: An earlier version of this question got very little attention, so I've edited it to explain an intended application and to clarify the question a bit. The latter edit involved replacing "discrete opfibration" (and notation $DopF$, $DopFS$) with "discrete fibration" (and notation $DF$, $DFS$) throughout, because it looked cleaner.

REPLY [3 votes]: In general, the answer is "No": the category $DFS_{C/}$ of surjective discrete fibrations under $C$ need not have a terminal object. This is due to the following:
Lemma. Let $C$ be the codiscrete category with $n$ objects. For any group $G$ with $n$ elements, there is a surjective discrete fibration $C \to \mathsf{B}G$ (where $\mathsf{B}G$ is the delooping of $G$).
Proof. The regular action of $G$ on itself by right multiplication gives rise to a functor $\mathsf{B}G^{op} \to \mathsf{Set}$ whose category of elements is isomorphic to  $C$. $\square$
When $n$ is composite, there are at least two non-isomorphic groups $G \not\cong H$ with $n$ elements, and these will give rise to non-isomorphic discrete fibrations $p \colon  C \to \mathsf{B} G$ and $q \colon C \to \mathsf{B} H$.
Further, any maps out of $p$ and $q$ in $DFS_{C/}$ will necessarily be isomorphisms (as noted by David's reply to Noam).
So if there were a terminal discrete fibration $t \colon C \to S$, we must have $\mathsf{B}G \cong S \cong \mathsf{B}H$, which contradicts our assumption that $G \not\cong H$.
But $p$ and $q$ above are minimal in some sense: all maps out of them are isomorphisms. So while $DFS_{C/}$ has no terminal objects, it does have multiple minimal ones.
(Still, any attempt to classify minimal objects in $DFS_{C/}$ for all (finite) categories $C$ must include the classification of all (finite) groups!)
However, if we are willing to restrict the categories $C$ that we want to consider, then we have:
Theorem. Let $C$ be such that for all $x \in C$, the slice $C/x$ has no non-trivial automorphisms. Then $DFS_{C/}$ has a terminal object.
I've written the proof in this draft. The idea is to construct a category $S$ with one object for each isomorphism class $[C/x]$ of slice categories in $C$, and then show that we have a discrete fibration $C \to S$.