Let $U : \mathbf{CRing} \to \mathbf{Set}$ be the forgetful functor. For any functor $F : \mathbf{CRing} \to \mathbf{Set}$ consider the class of natural transformations $$\mathcal{O}(F) := \mathrm{Hom}(F,U).$$ It is clear that $\mathcal{O}(F)$ carries the structure of a commutative "ring" (it doesn't have to be a set). When $F$ is representable, say $F \cong \mathrm{Hom}(S,-)$, then $ \mathcal{O}(F) \cong S$ by the Yoneda Lemma.

More generally, in functorial algebraic geometry, we can consider schemes as (special) functors $F : \mathbf{CRing} \to \mathbf{Set}$, and $\mathcal{O}(F) = \mathrm{Hom}(F,\mathbb{A}^1)$ is the ring of global sections of $F$. In particular, this is a set. It follows that this also holds for algebraic spaces.

Question. Is there a classification of those functors $F : \mathbf{CRing} \to \mathbf{Set}$ for which the class $\mathcal{O}(F)$ is actually a set?

If this is not solvable: Is there a classification of the continuous functors $F$ with the property? At the nlab page on total categories there is an example of a continuous functor $F$ which is not representable. If this is also not possible, I am looking for sufficient conditions. Although I wrote about classes, I am also very up for using universes instead. But I hope that the answer to my question does not depend on some subtle set theory axioms.

Edit. It is clear that $F$ being small is sufficient condition.

Let $U : \mathbf{CRing} \to \mathbf{Set}$ be the forgetful functor. For any functor $F : \mathbf{CRing} \to \mathbf{Set}$ consider the class of natural transformations $$\mathcal{O}(F) := \mathrm{Hom}(F,U).$$ It is clear that $\mathcal{O}(F)$ carries the structure of a commutative "ring" (it doesn't have to be a set). When $F$ is representable, say $F \cong \mathrm{Hom}(S,-)$, then $ \mathcal{O}(F) \cong S$ by the Yoneda Lemma.

More generally, in functorial algebraic geometry, we can consider schemes as (special) functors $F : \mathbf{CRing} \to \mathbf{Set}$, and $\mathcal{O}(F) = \mathrm{Hom}(F,\mathbb{A}^1)$ is the ring of global sections of $F$. In particular, this is a set. It follows that this also holds for algebraic spaces.

Question. Is there a classification of those functors $F : \mathbf{CRing} \to \mathbf{Set}$ for which the class $\mathcal{O}(F)$ is actually a set?

If this is not solvable: Is there a classification of the continuous functors $F$ with the property? At the nlab page on total categories there is an example of a continuous functor $F$ which is not representable. If this is also not possible, I am looking for sufficient conditions. Although I wrote about classes, I am also very up for using universes instead. But I hope that the answer to my question does not depend on some subtle set theory axioms.

Edit 1. It is clear that $F$ being small is sufficient condition.

Edit 2. Here is an example where $\mathcal{O}(F)$ is a class (similar to the nlab example). For every infinite cardinal $\lambda$ let $L_{\lambda}$ be a field of cardinality $\lambda$. When $R$ is a ring and $\lambda > \mathrm{card}(R)$, there can only be a homomorphism $L_{\lambda} \to R$ when $R=0$. Therefore, $F(R) := \coprod_{\lambda} \mathrm{Hom}(L_{\lambda},R) ~~/~~ \coprod_{\lambda} \mathrm{Hom}(0,R)$ is a set: for $R = 0$ we have $F(R)=\{\star\}$, and for $R \neq 0$ we have $F(R)=\coprod_{\lambda \leq \mathrm{card}(R)} \mathrm{Hom}(L_{\lambda},R)$. Notice that the quotient "does not bother" $\mathcal{O}$ since $U(0)=\{\star\}$, so we have $\mathcal{O}(F) = \prod_{\lambda} L_{\lambda}$, which is not a set.

Let $U : \mathbf{CRing} \to \mathbf{Set}$ be the forgetful functor. For any functor $F : \mathbf{CRing} \to \mathbf{Set}$ consider the class of natural transformations $$\mathcal{O}(F) := \mathrm{Hom}(F,U).$$ It is clear that $\mathcal{O}(F)$ carries the structure of a commutative "ring" (it doesn't have to be a set). When $F$ is representable, say $F \cong \mathrm{Hom}(S,-)$, then $ \mathcal{O}(F) \cong S$ by the Yoneda Lemma.

More generally, in functorial algebraic geometry, we can consider schemes as (special) functors $F : \mathbf{CRing} \to \mathbf{Set}$, and $\mathcal{O}(F) = \mathrm{Hom}(F,\mathbb{A}^1)$ is the ring of global sections of $F$. In particular, this is a set. It follows that this also holds for algebraic spaces.

Question. Is there a classification of those functors $F : \mathbf{CRing} \to \mathbf{Set}$ for which the class $\mathcal{O}(F)$ is actually a set?

If this is not solvable: Is there a classification of the continuous functors $F$ with the property? At the nlab page on total categories there is an example of a continuous functor $F$ which is not representable. If this is also not possible, I am looking for sufficient conditions. Although I wrote about classes, I am also very up for using universes instead. But I hope that the answer to my question does not depend on some subtle set theory axioms.

Let $U : \mathbf{CRing} \to \mathbf{Set}$ be the forgetful functor. For any functor $F : \mathbf{CRing} \to \mathbf{Set}$ consider the class of natural transformations $$\mathcal{O}(F) := \mathrm{Hom}(F,U).$$ It is clear that $\mathcal{O}(F)$ carries the structure of a commutative "ring" (it doesn't have to be a set). When $F$ is representable, say $F \cong \mathrm{Hom}(S,-)$, then $ \mathcal{O}(F) \cong S$ by the Yoneda Lemma.

More generally, in functorial algebraic geometry, we can consider schemes as (special) functors $F : \mathbf{CRing} \to \mathbf{Set}$, and $\mathcal{O}(F) = \mathrm{Hom}(F,\mathbb{A}^1)$ is the ring of global sections of $F$. In particular, this is a set. It follows that this also holds for algebraic spaces.

Question. Is there a classification of those functors $F : \mathbf{CRing} \to \mathbf{Set}$ for which the class $\mathcal{O}(F)$ is actually a set?

If this is not solvable: Is there a classification of the continuous functors $F$ with the property? At the nlab page on total categories there is an example of a continuous functor $F$ which is not representable. If this is also not possible, I am looking for sufficient conditions. Although I wrote about classes, I am also very up for using universes instead. But I hope that the answer to my question does not depend on some subtle set theory axioms.

Edit. It is clear that $F$ being small is sufficient condition.