Can $\mathsf{Set}$ be seen as a (non-trivial) 2-category? I know that $\mathsf{Rel}$ can be seen as a 2-category with inclusion of relations as 2-morphisms, when passing to $\mathsf{Set}$, relations become functions and inclusions are constrained to be "equalities" (see Is $\mathsf{Set}$ a 2-category?), is there any other way where we can see $\mathsf{Set}$ as a non-trivial 2-category? If not, how can we prove that it cannot be other than a trivial 2-category?
 A: There is a (not very interesting) construction which works for every category $\mathcal{C}$: Take any commutative monoid $M$. For parallel morphisms $f,g$ we define a $2$-morphism $f \to g$ to be some element of $M$ when $f=g$, and otherwise there is none. The composition of $2$-morphisms is the composition in $M$.
A: This should be a comment, but it's too long so there it is.
You could always define a new concept of arrows that probe deeper inside the inner structures of $Sets$. $Sets$ should actually be called "$Card$" (as category of cardinals). 
Knowing that elements of sets are sets themselves, its not difficult to come up with some concepts of $2$-functions $f: A \rightarrow B$ seen as a pair $(g, (g_a))$, where $g: A \rightarrow B$ is an usual function of set, and $(g_a)$ is a family of functions such that $g_a:a \rightarrow g(a)$ for all $a$ in $A$. These $2$-functions would play the role of $1$-arrows in your $2$-category  $2$-$Sets$. Now, you have to find a natural definition for $2$-arrows (i.e., maps of $2$-functions). It surely can be done (I didn't try), and you could even be able to make $Sets$ into an $\omega$-category, that would actually encodes all the membership tree (thus being the "true" category of set theory).
You have to take care of empty set I think, but I guess it can be done.
edit: Let $A$ and $B$ be two sets, let us write a 2-map between $A$ and $B$ as $(f, f_a)$ with $f: A \rightarrow B$ and for each $a$ in $A$, a map $f_a: a \rightarrow f(a)$.
You can define a composition here as $(g, g_b) \circ (f, f_a) = (g \circ f, (g\circ f)_a = g_{f(a)} \circ f_a)$. This composition is associative, and identity is obviously given by $(Id_A, ({Id_A})_a = Id_a)$, turning the collection of sets where $1$-arrows are $2$-maps into a $1$-category. We now have to define $2$-arrows and show that the composition is functorial. Let us define a $2$-arrow $\eta: (f,f_a) \Rightarrow (g,g_a)$ as a collection of maps $\eta_a: f(a) \rightarrow g(a)$ for each $a$ in $A$, such that $\eta_a \circ f_a = g_a$. This is highly inspired from the concept of natural transformation as you might see. I'm going to check if this turn the above category as a $2$-cat (it should). The vertical composition is simply given by $(\xi \circ_1 \eta)_a = \xi_a \circ \eta_a$, making the collection of arrows between $2$-maps between two fixed sets a $1$-cat as it should be. It is left to define an horizontal composition and verify the functoriality.
Note that the initial element is given by the empty set and that if $(f,f_a)$ is a $2$-map between $A$ and $B$ with empty set in the image of $f$, then the empty set is necessarily an element of $A$ with $f(\emptyset) = \emptyset$.
