Unique internal functor between internal categories on $A$ and $B$ with a given functor $F:A\to B$ on internal objects Given a category $A$, there is a natural internal category structure $(A^2,A,i_A,\circ_A)$ on it. here $2$ represents the category with only two object $0$ and $1$, and only one non-identity morphism $0\to 1$. Therefore, a functor $2\to A$ represents an arrow in $A$.
A functor $F:A\to B$ clearly induces a internal functor between the internal categories $(A^2,A,i_A,\circ_A)$ and $(B^2,B,i_B,\circ_B)$. The internal functor is given by the pair ($F: A\to B$ , $F^2:A^2\to B^2$), where $F^2$ sends objects of $A^2$ to the object in $B^2$ which is a transpose of the arrow $2\to B$ obtained by applying $F$ to an arrow represented as a functor $2\to A$.
Now given a pair $(F:A\to B,G:A^2\to B^2)$, can we use the definition of an internal functor to prove that $G$ must be $F^2$?
It is actually Theorem 24 in McLarty's "Axiomatizing a category of categories".
Here is a link to the paper: https://www.jstor.org/stable/2275472?casa_token=hi4gsrP5DHMAAAAA:l5929tBx2JWhyJS9B1aoR8iB3AaD1o8NByGOmoj2MHTMoI-LrnB4EzfRSa-si01RyxIz4dHbuldXesD-x9LvW-TmmojxoTMS7XbUrNWEsykCsx0p
Although there is some special setting in the paper (He uses a functor from a triangle-shaped category for composition within a category), I tend to believe such a result holds even without the settings in the paper. The paper only writes out which axioms are used and I am stuck on proving this, thank you if you take your time to read the paper and give me some hints or complete proof using the terminology there. But without even looking into the paper, thank you for any idea on this!
 A: Yes.
Let's briefly recall what we have at our disposal. An internal category has an object $M$ of morphisms, an object $B$ of objects, and morphisms $s,t:M\to B$, $\eta: B\to M$, and $\mu: M\times_B M\to M$ subject to some axioms. An internal functor between internal categories $(B,M)$ and $(B', M')$ is a pair of morphisms $B\to B'$ and $M\to M'$ that commute with all of the structural arrows.
We also know the structural morphisms of the internal categories in question. Namely for $A$ a category, the internal category has $M_A=A^2$, $B_A=A$, $s_A$ and $t_A$ induced by the functors $s,t:*\to 2$, $\eta_A$ induced by the terminal map $2\to *$, and the multiplication $\mu_A$ is induced by the inclusion of the composite morphism into the category $3=2\amalg_* 2$.
Now suppose that $(F,G)$ is a morphism from the internal category $(A,A^2)$ to the internal category $(B,B^2)$. If $f:a\to a'$ is an object in $A^2$, then $Gf$ (an object of $B^2$) has source $Fa$ and target $Fa'$ since $G$ respects sources and targets. I'll write $Gf:Fa\to Fa'$ for the morphism in $B$ as well.
Now, consider the following morphism in $A^2$, which I'll take as oriented vertically, so the source (as an object of $A^2$) is at the top of the square, target at the bottom and the $s_A$ functor takes the square to its left side and the $t_A$ functor takes the square to its right side.
$$
\require{AMScd}
\begin{CD}
a @>1_a>> a\\
@V1_aVV @VVfV\\
a @>>f> a'\\
\end{CD}
$$
What do we get if we apply $G$ to these morphisms? Well, the left and right side must agree with $F$, since $G$ respects $s_A$ and $t_A$, and $G$ on the identities (as objects in $A^2$) also agrees with $F$, since $1_a=\eta a$, thus $G1_a = G\eta a = \eta Fa=1_{Fa}$ as objects in $B^2$. Therefore we get
$$
\begin{CD}
Fa @>1_{Fa}>> Fa\\
@V1_{Fa} VV @VVFf V\\
Fa @>>Gf> Fa'\\
\end{CD}.
$$
Since this commutes, $Ff=Gf$. Now, this only seems to show that $G$ agrees with $F^2$ on the objects of $A^2$. Maybe $G$ and $F^2$ differ on morphisms. However, if we take an arbitrary morphism of $A^2$ viewed as a commutative square in the same way as before, we already know $G$ and $F^2$ agree on the sides of the square, since $G$ respects sources and targets, and since $G$ and $F^2$ agree on objects of $A^2$, they also agree on the top and bottom of the square as well. Thus $G$ and $F^2$ agree on morphisms.
