If I have a functor $D_n : \mathsf{Cat} \to \mathsf{Set}$ that maps a category into the set of all $n$-tuples of composable morphisms, $D_n(C) = A_1 \to A_2 \to A_3 \to \dots \to A_n$, what would its representing object be?
I have a gut feeling it would be a category with a single object and $n$ endomorphisms? But I also thought of it being a thin ordered set with $n$ objects. Either way, I don't know how to proceed or how I would prove it once I find an answer.
Consider the category $1\xrightarrow{t_1} 2\rightarrow\dots\xrightarrow{t_{n-1}} n$ with $n$ objects and $n-1$ composable morphisms. We want to show that there exist a natural bijection $D_n(\mathcal{C})\cong Hom_{\mathcal{Cat}}(1\rightarrow 2\rightarrow\dots\rightarrow n,\mathcal{C})$ for any category $\mathcal{C}$.
Let $\textbf{A} \ : \ A_1 \xrightarrow{f_1} A_2 \rightarrow\dots \xrightarrow{f_{n-1}} A_n$ be an $n$-tuple of composable morphisms in $\mathcal{C}$. Consider the functor $F_\textbf{A}$ from the indexed category $1\rightarrow 2\rightarrow\dots\rightarrow n$ to $\mathcal{C}$, defined on objects as $i\mapsto A_i$, and on morphisms as $t_i \mapsto f_i$ for any $i=1,\dots,n-1$. One can show that the function $D_n(\mathcal{C})\rightarrow Hom_{\mathcal{Cat}}(1\rightarrow 2\rightarrow\dots\rightarrow n,\mathcal{C})$ sending an $n$-tuple $\textbf{A}$ of composable morphisms to the functor $F_\textbf{A}$ defined before, is a bijection.
Moreover, given a functor $F:\mathcal{C}\rightarrow\mathcal{D}$, the following diagrams commutes $$ \require{AMScd} \begin{CD} D_n(\mathcal{C}) @> >> Hom_{\mathcal{Cat}}(1\rightarrow 2 \rightarrow\dots\rightarrow n,\mathcal{C})\\ @V {D_n(F)} VV @VV {F {\circ} -} V \\ D_n(\mathcal{D}) @> >> Hom_{\mathcal{Cat}}(1\rightarrow 2 \rightarrow\dots\rightarrow n,\mathcal{D})\\ \end{CD} $$ Thus, we've showed the bijection $D_n(\mathcal{C})\cong Hom_{\mathcal{Cat}}(1\rightarrow 2\rightarrow\dots\rightarrow n,\mathcal{C})$ is natural, and so the functor $D_n$ is represented by the category $1\rightarrow 2\rightarrow\dots\rightarrow n.$