Let $\mathcal{C}$ be a category. By diagram I mean a covariant functor $F\colon\mathcal{J}\to\mathcal{C}$ for some category $\mathcal{J}$. In this source it is said that a commutative diagram is a diagram $F\colon\mathcal{J}\to\mathcal{C}$ such that there exists a posetal category $\mathcal{P}$ and diagrams $G\colon\mathcal{J}\to\mathcal{P}$ and $H\colon\mathcal{P}\to\mathcal{C}$ such that the composition functor $H\circ G$ is naturally isomorphic to $F$.
My question is: can we drop the condition that $H\circ G$ must be naturally isomorphic to $F$ and just instead insist that $H\circ G=F$? Would such a definition be just as general as the quoted definition? It appears to me that if $H\circ G$ is naturally isomorphic to $F$, then I should be able to modify my functor $H$ to some functor $\tilde{H}\colon\mathcal{P}\to\mathcal{C}$ in such a way that $\tilde{H}\circ G=F$. Is this the case?
I think the issue is that we want to factor through a partially ordered set, which in particular entails the antisymmetry condition ($x\leq y \land y\leq x \implies x=y$).
Consider as diagram shape $\mathcal{I}$ the free living isomorphism $\mathcal{I}=(0\cong 1)$, where $\mathcal{I}(i,j)=\{\ast\}$ for all $i,j \in \{0,1\}$. Consider a functor $F:\mathcal{I}\rightarrow\mathcal{C}$, which picks two distinct (but isomorphic) objects. For example something like $\Bbb Z/2$ and $\{\pm 1\}$ as abelian groups. You will never be able to factor this functor through a single poset (this poset has necessarily to be the terminal category $\mathbb{1}$, which has only one object!). This deficit is resolved by either requiring a natural isomorphism to our original functor or allowing for non-automorphic isomorphisms in our poset structure. In your link this is called a proset, other words for that are posetal category or thin category.