\text{Given } H:\mathcal{C} ^{\mathrm{op}}{\texttimes} \mathcal{C}\to \mathrm{Set}
$$ $H$ is a *profunctor*.
Then $$
\int_{X:\mathcal{C} } H(X,X)=\{ \alpha :\prod_{X:\mathcal{C} }. \forall f:\mathcal{C} [A,B]. H(A,f) \alpha_A = H(f,B) \alpha _B \}
$$Is exactly a [[natural transformation]].
Note that this is a set because we can define equality.
A profunctor is generalization of relations and [[bimodule]].