If the objects of a category form a proper class, do the arrows necessarily form a proper class too? In some categories, like $\text{Set}$ or $\text{Group}$, the objects are "constructed" out of sets (or are sets, possibly with additional structure). In order to avoid paradoxes, the collection of objects is therefore a proper class✱.
If the class of objects is proper, is it possible for the collection of arrows to still be a set?
A not-convincing argument against it is that each object gets an identity arrow therefore there are "too many" arrows to be a set. I don't trust intuitions about size with things as big as $\text{Set}$ though.
A not-convincing argument that it's plausible is that arrows are opaque and I, the category-maker, get to freely pick the labels for the arrows, the definition of the composition relation, and the source and destination for each of the arrow-labels. How do I show that I can't come up with a set of labels big enough to label all of my arrows? 

✱ I don't know whether it makes sense to start with a "possibly-proper class of all sets satisfying some predicate" and then inspect the class in some way to see if it's a proper class or not.
 A: They have to form a proper class as well. 
For every object there is an identity arrow, so there must be more (or, in a discrete category, equally many) arrows than objects. 
A: It is possible to have a category whose objects are a proper class and yet the arrows form set, in fact they can form a singleton. Consider for example the category $\mathcal{C}$ whose objects is the class of all sets. We let $\hom(a,b) = \emptyset$ unless $a = b$ in which case we let $\hom(a,a) = \{ * \}$ for some fixed set $*$. We furthermore define composition via $* \circ * := *$. This is a category with a proper class of objects and only one arrow.

Let me add the following: If you are only interested in categories from the point of view as a category theorist (and not as a set theorist), you can, without loss of generality, assume that for any given category $\mathcal{C}$ we have $a \neq b \implies \mathrm{id}_{a} \neq \mathrm{id}_{b}$ and here's why:
Given $\mathcal{C}$ construct a new category $\mathcal{D}$ with the same objects as $\mathcal{C}$ and such that, for $a,b \in \mathrm{ob}(\mathcal{D})$
$$
\hom_{\mathcal{D}}(a,b) := \{ (a,b,f) \mid f \in \hom_{\mathcal{C}}(a,b) \}
$$
and
$$
(b,c,f) \circ_{\mathcal{D}} (a,b,g) := (a,c, f \circ_{\mathcal{C}} g).
$$
$\mathcal{C}$ and $\mathcal{D}$ are equivalent from the point of view of category theory and moreover they preserve all sorts of nice set theoretical properties (e.g. $a \mapsto \mathrm{id}^{\mathcal{C}}_{a}$ is definable if and only if $a \mapsto \mathrm{id}^{\mathcal{D}}_{a}$ is definable).
And $\mathcal{D}$ satisfies that $a \neq b \implies \mathrm{id}^{\mathcal{D}}_{a} \neq \mathrm{id}^{\mathcal{D}}_{a}$.
