The category of finite sets has homotopy cardinality $e$, because $$ |{\bf FinSet}|=\sum_{n=0}^{\infty}\frac{1}{\left|\operatorname{Aut}\ [n]\right|}=\sum_{n=0}^{\infty}\frac{1}{n!}. $$
What is the homotopy cardinality of ${\bf FinCat}$, the category of finite categories?
Sets are $0$-categories. According to the periodic table of categories, we have $$ |{\bf FinCat_{-2}}|=1,\qquad |{\bf FinCat_{-1}}|=2,\qquad |{\bf FinCat_{0}}|=e. $$ Is there a reason to expect this to be an increasing sequence?
There are infinitely many inequivalent finite categories with no automorphisms, so the homotopy cardinality is infinite. For instance, if $G=(V,E)$ is any finite graph, we can consider the set $V\sqcup E$ to be partially ordered by saying an edge is greater than both of its vertices. Clearly the graph can be recovered from this poset, so every automorphism of the poset gives an automorphism of the graph. But there are infinitely many finite graphs with no nontrivial automorphisms (in fact, the probability that a random finite graph has no nontrivial automorphisms goes to $1$ as the number of vertices goes to $\infty$). So we get infinitely many finite posets with no nontrivial automorphisms.