Is the collection of all subfunctors of a given representable endofunctor of $\textbf{Set}$ a set? Let $F$ be a representable endofunctor of the category Set of sets. 
(1) Is the collection of all subfunctors of $F$ a set?
(2) Same question with "representable endofunctor of Set" replaced by "representable functor from $\mathcal C$ to Set, where $\mathcal C$ is a locally small category". 
(3) Same question with "subfunctors" replaced by "quotients".
[Of course, (2) is a generalization of (1).]
Edit. If we drop the representability assumption, we can find examples where the collections of subfunctors and of quotients are not sets. Indeed, in the notation of 
Peter Freyd, Ross Street, On the size of categories, TAC 1 (1995) pp.174-185 http://www.tac.mta.ca/tac/volumes/1995/n9/v1n9.pdf
let $\mathcal A$ be the category of sets, and, for each set $K$, let $U_K$ be the image of the endomorphism $\theta K$ of $T$. Then $U_K$ is a quotient and a subfunctor of $T$, and $U_K\neq U_L$ for $K$ and $L$ of different cardinalities. (Clearly $T$ is not representable.)
 A: For (1), note that if $G$ is a subfunctor of $\text{Hom}(X,-)$ and if $\alpha\in G(Y)$ for some set $Y$, and $\beta:X\to Z$ induces the same equivalence relation on $X$ as $\alpha$ does (i.e., if $\alpha(x)=\alpha(x')\Leftrightarrow\beta(x)=\beta(x')$ for $x,x'\in X$), then $\beta$ factors through $\alpha$, and so $\beta\in G(Z)$.
So $G$ is determined by the class of equivalence relations on $X$ induced by elements of $G(Y)$ for sets $Y$. But there is only a set of such classes.
For (2), take $\mathcal{C}$ to have objects $X$ and $\{Y_i\vert i\in I\}$ for some proper class $I$, with the only non-identity morphisms being a morphism $X\to Y_i$ for each $i$. Then $\text{Hom}(X,-)$ has a proper class of subfunctors, since $\text{Hom}(Y_i,-)$ is isomorphic to a subfunctor for every $i\in I$.
For (3), note that if $G$ is a quotient functor of $\text{Hom}(X,-)$, and $\alpha,\beta\in\text{Hom}(X,Y)$, then $\alpha=\beta$ in $G(Y)$ if and only if $\bar{\alpha}=\bar{\beta}$ in $G\left(\text{im}(\alpha)\cup\text{im}(\beta)\right)$, where $\bar{\alpha}$ and $\bar{\beta}$ are the functions obtained by restricting the codomain of $\alpha$ and $\beta$. This is because if $r:Y\to\text{im}(\alpha)\cup\text{im}(\beta)$ is any retraction, then $r\circ\alpha=\bar{\alpha}$ and $r\circ\beta=\bar{\beta}$.
So $G$ is determined by the choice of a set of pairs of functions from $X$ to sets with cardinality at most twice that of $X$, and there is only a set of such choices up to isomorphism.
