If an arrow category is like a path what's like a cylinder? The arrow category is isomorphic to a functor category out of the interval category.
$$\mathrm{Arr}( C) = [\mathrm I, C]$$
It is like a very simple/degenerate case of a path object.
A "loop" speaking loosely would just be an arrow (possibly identity?) with equal source and target.
You would expect it to have an adjoint something like
$$? = C \times \mathrm{I}$$
And a sort of "suspension" category would be some sort of quoitent or pushout that I don't quite understand equating (x, 0) and (y, 1).
But I can't really figure out how to conceptualize these as concrete things. What does it mean to be an object of a "cylinder" category?
Maybe you could think of it as mapping between source/targets?
 A: The way I like to think about $\mathcal C \times I$ is that it classifies natural transformations.
Just like $X \times I$ classifies homotopies for topological spaces $X$.
Natural transformations $\phi:\mathcal F \rightarrow \mathcal G$ between two functors $\mathcal C \rightarrow \mathcal D$ are in bijection with functors $H$ that fit into the diagram.

We simply set $H((c,0) \rightarrow (c,1)) = \phi_c$ to establish this bijection.
A: This is more a lengthy comment than an answer, but I hope it is relevant enough to not be annoying...
As you noted the free living arrow $[1]=0\rightarrow 1$ gives rise to the arrow category of a category $\mathscr{C}$ via the functor category $\operatorname{Fun}([1],\mathscr{C})$. Similarly any totally ordered set $[n]=0\rightarrow 1 \rightarrow ... \rightarrow n$ gives rise to some sort of a category of paths/walks of length $n$ in $\mathscr{C}$ via $\operatorname{Fun}([n],\mathscr{C})$.
The category of categories $\mathsf{Cat}$ is cartesian closed, so we have an adjunction
$$-\times [n] \dashv \operatorname{Fun}([n], -).$$
The product $\mathscr{C}\times [n]$ can be computed to consist of $n+1$ copies of $\mathscr{C}$ with a fixed order and consecutive copies joined together. So in some sense it does constitute a cylinder.
The problem with this constructions is the following though: it is good practice to have different path-spaces respectively different cylinders to be at least weakly equivalent. But in this case there is (to my knowledge) no good notion of weak equivalence of categories making $\operatorname{Fun}([1],\mathscr{C}) \sim \operatorname{Fun}([n],\mathscr{C})$ true. The issue is with the directedness and different length of the categories $[n]$.
We can remedy the situation by considering the free living iso $I=0\overset{\rightarrow}{\underset{\leftarrow}{}}1$ and any chain $I_n$ of $n$ copies of it. Then everything mentioned works out analogously with the difference that eg. $\mathscr{C}\times I$ and $\mathscr{C}\times I_n$ are equivalent as categories.
This is not just a nice little fact. A homotopy equivalence with respect to the interval $I$ in $\mathsf{Cat}$ is precisely an equivalence of categories. In fact it gives a whole homotopy theory of categories (you may want to look up the folk/categorical model structure). As I understand it, developing category theory out of this perspective constitutes Prof. Cisinskis approach to $\infty$-categories.
I haven’t seen the construction of say unreduced suspension be written out or used anywhere yet. But it might be worth to have a look in the case of the interval being the free living iso.
A: I am not sure what the question is, but the "cylinder category" $C \times I$ is just the usual product of categories of $C$ and $I$, and indeed we have $\mathrm{Hom}(C \times I,D) \cong \mathrm{Hom}(C,\mathrm{Hom}(I,D))$. So this product category is already what you are looking for. The geometric realization $|-| : \mathbf{Cat} \to \mathbf{Top}$ (when done with a convenient category of topological spaces) commutes with finite products, so indeed $|C \times I| = |C| \times [0,1]$ is a cylinder in the topological sense.
You also asked for a "suspension category" $S(C)$ which should be a quotient of $C \times I$ in which $(x,0) \cong (x',0)$ and $(x,1) \cong (x',1)$ for all $x,x' \in C$, and such that the geometric realization is $|S(C)| = S(|C|)$. In fact, we might just adjoin these isomorphisms to $C \times I$, and keep the same objects. Any morphism $x \to x'$ in $C$ already induces a morphism $(x,0) \to (x',0)$, and we identify it with the adjoined isomorphism (similarly for $1$).
