In the "category of categories" $Cat$ (see e.g. here for a rigorous definition), there is a category $A$ with two objects and exactly one non-identity morphism. (Cf. "two-valued object", "bi-pointed object", and "cospan".)
Specifically, given the terminal category $\mathbf{1}$, there are exactly two distinct functors $\mathbf{1} \to A$, call them $s$ ("source") and $t$ ("target"), which of course correspond to the objects of $A$, and then the unique non-identity morphism within $A$ is denoted $s \overset{a}{\rightarrow} t$ ($a$ for "arrow").
Note that for any category $\mathcal{C}$, every morphism in $\mathcal{C}$ corresponds to a functor $A \to \mathcal{C}$. Moreover, building on the Cartesian closed structure of $Cat$, for any category $\mathcal{C}$ the exponential object/category $\mathcal{C}^A$ exists, and is usually called the "arrow category of $\mathcal{C}$ and denoted $Arr(\mathcal{C})$ or $\mathcal{C}^{\rightarrow}$ (cf. here, here, or here).
The category $A$ is sometimes denoted $\mathbf{2}$, although this risks confusion with the coproduct $\mathbf{1} \oplus \mathbf{1}$. Likewise, I have also seen it referred to as the "interval category", "the ordinal", or the "$1$-simplex", although all of those terms are ambiguous in the sense that they can (and more often do) refer to distinct notions than "the" category $A$.
Questions:
Does the category $A$ admit a characterization in terms of a universal property inside of $Cat$, allowing one to attempt to define analogous "arrow objects" in any category, e.g. $Set$ or $Top$? If so, is it a (co)limit, or another universal construction? Are there references describing this?
Does the category $A$ (defined in terms of such a universal property) possess analogues in other categories? Again, are there references describing this?
Attempt:
1. I claim that the morphism $a$ can be identified with the endofunctor $\alpha: A \to A$ such that (abusing notation by identifying objects with functors from $\mathbf{1}$) $\alpha(s) = t$, $\alpha(t)=t$, and sending all morphisms to $id_t$.
The category $A$ can be written using the diagram (also if anyone can format this better please edit the post to do so) $$\begin{matrix} A & \overset{\alpha}{\longrightarrow} & A \\ s \nwarrow & & \nearrow t \\ & \mathbf{1} & \end{matrix} $$ I then claim that $(A, s, t, \alpha)$ is universal in the following sense, namely given any other category $\mathcal{C}$ with two objects $C_1$ and $C_2$ with an endofunctor $F: \mathcal{C} \to \mathcal{C}$ that sends $C_1$ to $C_2$, i.e. $F(C_1) = C_2$, as in the following diagram: $$\begin{matrix} \mathcal{C} & \overset{F}{\longrightarrow} & \mathcal{C} \\ C_1 \nwarrow & & \nearrow C_2 \\ & \mathbf{1} & \end{matrix} $$ there is a unique functor $I: A \to \mathcal{C}$ such that $I \circ \alpha = F \circ I$ and $I \circ s = C_1$, namely the following commutes: $$\begin{matrix} A & & \overset{\alpha}{\longrightarrow} & & A \\ \vdots&s \nwarrow & & \nearrow t & \vdots \\ \exists! I & & \mathbf{1} & & \exists! I \\ \downarrow & C_1 \swarrow & & \searrow C_2 &\downarrow \\ \mathcal{C} & & \overset{F}{\longrightarrow} & & \mathcal{C} \end{matrix}$$ If so, this would make $A$ some "initial-like" object, even if not exactly a colimit.
Possible problems with attempted solution for 1.
This doesn't seem quite right because it might be conflating endofunctors and morphisms too much (but we can't compose a morphism inside of a category with a functor between categories in a commutative diagram, right?) and also, if there are multiple morphisms $C_1 \to C_2$ inside of $\mathcal{C}$, then there should be multiple functors $A \to \mathcal{C}$ satisfying corresponding commutativity requirements, one for each morphism? I don't see how the above (if it even exists) accomplishes that.
I don't see how we can "replace" the morphism $a$ with the endofunctor $\alpha$. The existence of $a$ "inside of" $A$ does not seem necessary for $\alpha$ to be well-defined as a functor.
Given $\mathbf{1} \oplus \mathbf{1}$, you could just consider the (composition of) functor(s) $\mathbf{1} \oplus \mathbf{1} \to \mathbf{1} \overset{t}{\to} \mathbf{1} \oplus \mathbf{1}$ which behaves exactly the same on all of the objects and morphisms that $A$ has in common with $\mathbf{1} \oplus \mathbf{1}$.
Is there any better choice of endofunctor that could be used to "replace" $a$?
2. Inside of $Set$, consider the two element set $\{ \ast, \dagger \}$ (thus equipped with morphisms from the one-element set targeting both $\ast$ and $\dagger$ as well as with an endomorphism $f$ such that $f(\ast) = \dagger$ and $f(\dagger) = \dagger$. Then replacing $\alpha$ with $f$, $A$ with $\{\ast, \dagger\}$, $s$ with $\ast$, and $t$ with $\dagger$ (abusing notation and conflating elements of a set with maps from the one-element set to the set), this behaves the same way with respect to commutative diagrams as $A$ is supposed to in $Cat$ above.
Note on 2-category theory:
During internet searches I have found several references defining constructions similar to $A$ using $2$-categories. There is also an answer to a related Math.SE question which, when not defining $A$ in terms of itself, uses $2$-categories (via lax colimits). However, I do not think that any characterization of such a simple structure as $A$ via the framework of $2$-categories is or can be useful. I also don't see why it should be necessary.
If $2$-category theory really is necessary to axiomatize $A$ (which still seems doubtful and overkill to me), is the reason why in order to be able to capture the "internal structure" of $A$? (I.e. vis a vis the morphism $a$ and being able to distinguish $A$ from the category $\mathbf{1} \oplus \mathbf{1}$ which lacks $a$)?
And if so, can you provide a proof, or a reference proving, that $A$ can not be axiomatized using only $1$-category theory, and that $2$-category theory is truly necessary? In particular, is there some sense in which it is impossible to distinguish $\mathbf{1} \oplus \mathbf{1}$ and $A$ only using functors, and the full $2$-category structure of $Cat$ using natural transformations is somehow necessary to distinguish them?
