The existence of a functor between the category of all sets Let $\mathsf{Set}$ be the category consisting of all non-empty sets and its morphisms are mappings.
If $F$ is a functor from $\mathsf{Set}$ to $\mathsf{Set}$, then $\{F\}$ is a set. Consequently, $F(\{F\})$ is also a set. But this is very similar to the circular definition of Russell's paradox.
Of coure, if $F$ is a functor between two small categories, $F$ must be well-defined. But I don't know how to explain this kind of situation in $\mathsf{Set}$.
 A: You're right to be worried about this! However the way we sidestep Russell's paradox is entirely analogous to how we'll sidestep this issue. Let $V$ be the universe of all sets. Recall the object $\{ x \in V \mid \varphi(x) \}$ is, in general, not a set! It is a bigger object called a class, and only certain (intuitively, "small") classes are sets. Then if $R = \{ x  \in V \mid x \not \in x \}$, it doesn't even make sense to ask if $R \in R$, because $R$ is not a set!
For similar reasons, $F$ isn't a set either! If $F$ is a functor from $\mathsf{Set}$ to itself, then its domain is $V$. Recall in set theory we identify a function with its graph, so formally we have
$F = \{ (x, Fx) \mid x \in V \}$, which is as big as $V$! So $F$ is a proper class, and $\{ F \}$ isn't something we can really write down.
There are a few ways we can talk about $F$ even though it isn't a set1. One common approach is to fix a large cardinal $\kappa$ and say $\mathsf{Set}$ is actually the category of "small sets". That is, sets of size $< \kappa$. Now we see $\mathsf{Set}$ does have only a set worth of objects, and so we can write down $F$ as a set too (but now $F$ isn't small, and we've cleverly sidestepped the issue again!).
1: Provided we insist upon trying to formalize category theory in $\mathsf{ZFC}$ or something similar, which some people don't do.

I hope this helps ^_^
