Can an algebraic variety be described as a category, in the same way as a group?

Can an algebraic variety be described as a category, in the same way as a group? A group can be considered a category with one object, with elements of the group the morphisms on the object.

The conjectural remarks by Qiaochu are supported by the following theorem (joint work with Alexandru Chirvasitu):

Let $X,Y$ be schemes, where $X$ is quasi-compact and quasi-separated. Then $f \mapsto f^*$ establishes an equivalence between $\hom(Y,X)$ and the category of cocontinuous symmetric monoidal functors $\mathsf{Qcoh}(X) \to \mathsf{Qcoh}(Y)$.

Therefore, the category of (nice) schemes embeds fully faithfully into the $2$-category of cocomplete symmetric monoidal categories. Large parts of algebraic geometry can be translated and generalized by means of this embedding; this will be hopefully explained in full detail in my thesis. For example, under suitable finiteness assumptions, one can construct the "projective tensor bundle" $\mathbb{P}_{\otimes}(\mathcal{E})$ for an object $\mathcal{E}$ of a cocomplete symmetric monoidal category $\mathcal{C}$, which classifies invertible quotients of $\mathcal{E}$, and when $\mathcal{C}=\mathsf{Qcoh}(S)$ for some scheme $S$ this coincides with $\mathsf{Qcoh}(\mathbb{P}_S(\mathcal{E}))$.

In this picture, the monoidal structure is crucial; there are lots of auto-equivalences of $\mathsf{Qcoh}(X)$ which are not induced by automorphisms of $X$. On the other hand, if $X,Y$ are quasi-separated schemes such that $\mathsf{Qcoh}(X)$ and $\mathsf{Qcoh}(Y)$ are equivalent as abstract categories, then $X$ and $Y$ are isomorphic; this is a quite deep theorem (proved by Gabriel for noetherien schemes, then generalized by Rosenberg, and corrected by Gabber) and has motivated a new perspective on non-commutative algebraic geometry using abelian categories (see the work by Artin, Zhang, Rosenberg and others).

The answer is yes, in a sense, but it looks nothing like the result for groups. The easy version of the result is for affine varieties and says the following. Fix an algebraically closed field $k$.

Theorem ("Gabriel-Rosenberg reconstruction"): An affine variety $X$ can be reconstructed from the ($k$-linear) abelian category of modules over the ring of regular functions $k[X]$.

The idea of the proof is that we can recover $k[X]$ (as a $k$-algebra) by a general categorical construction called the center of a category; see this blog post for details. There is a harder version of this theorem for not-necessarily-affine not-necessarily-varieties in which the category of modules is replaced by the category of quasicoherent sheaves on $X$. I don't know anything about this.

In a previous version of this answer I tried to promote the assignment $X \mapsto k[X]\text{-Mod}$ to a functor in such a way that it was faithful and full, but as it turns out the natural way to do this seems to require that it be a contravariant functor rather than a covariant functor. I haven't worked out the details here; part of these slides briefly discuss the story. (I think the theorem of Lurie cited there is Theorem 5.11 in Tannaka Duality for Geometric Stacks.)

The issue is the following. If $R, S$ are two rings and $f : R \to S$ is a morphism, $f$ induces two functors induced between their categories of modules. One of them might be called "restriction"

$$S\text{-Mod} \to R\text{-Mod}$$

and it is obtained by regarding an $S$-module as an $R$-module with $R$ acting via $f$. The other might be called "induction"

$$R\text{-Mod} \to S\text{-Mod}$$

and it is obtained by tensoring with $S$ as an $(S, R)$-bimodule. The two functors are adjoints. When $R, S$ are commutative, induction appears to be the more geometrically natural functor whereas I was trying to do things using restriction. Induction satisfies various nice properties and one might hope to single out those functors $R\text{-Mod} \to S\text{-Mod}$ arising from a morphism $R \to S$ by some categorical property. This appears to be doable when $R, S$ are commutative but I haven't worked through the details.