Can we sketch $\mathbf{Set}^{\mathrm{op}}$? Is there a large limit sketch whose category of models is equivalent to $\mathbf{Set}^{\mathrm{op}}$?
Is there perhaps even a large limit sketch $\mathscr{S}$ generated by powers of an object $X$ such that the forgetful functor $\mathbf{Mod}(\mathscr{S}) \to \mathbf{Set}$ is equivalent to $\mathrm{Hom}(-,2) : \mathbf{Set}^{\mathrm{op}} \to \mathbf{Set}$?
(There cannot be a small limit sketch because $\mathbf{Set}^{\mathrm{op}}$ is not locally presentable.)
 A: A fundamental theorem of elementary topos theory is that the contravariant powerobject functor $\Omega^{(-)} : \mathcal{E}^\textrm{op} \to \mathcal{E}$ is monadic for any elementary topos $\mathcal{E}$.
In particular, $2^{(-)} : \textbf{Set}^\textrm{op} \to \textbf{Set}$ is monadic.
It is also known that if $U : \mathcal{A} \to \textbf{Set}$ is monadic, then $\mathcal{A}$ is equivalent to the category of models of a (large infinitary) Lawvere theory $\mathcal{T}$, namely the opposite of the Kleisli category of the monad, hence also equivalent to the category of models of a (large infinitary) product sketch.
Let me be more precise about the second part.
We may assume without loss of generality that $U : \mathcal{A} \to \textbf{Set}$ is the forgetful functor from the Eilenberg–Moore category of a monad $(T, \eta, \mu)$ on $\textbf{Set}$.
Given an algebra $(A, \alpha)$, the functor $\mathcal{A} ({-}, (A, \alpha))$ defines a product-preserving functor $\mathcal{T} \to \textbf{Set}$, i.e. a model of $\mathcal{T}$, and it is easy to check that we get a fully faithful embedding $\mathcal{A} \to \textbf{Mod} (\mathcal{T})$.
On the other hand, suppose we have a product-preserving functor $M : \mathcal{T} \to \textbf{Set}$.
Let $F^{(-)} : \textbf{Set}^\textrm{op} \to \mathcal{T}$ be (the opposite of) the free algebra functor.
Then $M (F^{(-)}) : \textbf{Set}^\textrm{op} \to \textbf{Set}$ preserves products, hence is represented by some set $A$ and some $m \in M (F^A)$.
We may assume w.l.o.g. that $A = M (F^1)$.
Let $f \in T A$.
This determines a morphism $F^A \to F^1$ in $\mathcal{T}$, hence induces a map $f : M (F^A) \to M (F^1)$.
Define $\alpha : T A \to A$ by $\alpha (f) = f (m)$.
You can then verify that $(A, \alpha)$ is an algebra and that $M$ is isomorphic to the restriction of the functor represented by $(A, \alpha)$.
Thus $\mathcal{A} \to \textbf{Mod} (\mathcal{T})$ is also essentially surjective on objects, so we indeed have an equivalence of categories.
