Your problem is that your intuitions about the empty set don't fit the mathematical definitions; where functions etc. end up existing vacuously. In this answer, given your background I'll discuss the category Set. This category has as it's objects sets and as its morphisms functions between sets.
In this setting, your question asks what the morphisms with domain $\emptyset$ are. The point is that the definition of a function $f:A \to B$ is "for each $a \in A$ there is a unique $b \in B$ such that $f(a) = b$" (In set theoretic language $f:A \to B$ is actually a subset of $A \times B$ such that for each $a \in A$, there is a unique $b \in B$ such that $(a,b) \in f$).
In particular, this definition means that for any set $B$ there is a unique function from $\emptyset$ to $B$; namely $\emptyset$ itself! (though in the category we distinguish between the object $\emptyset$ and each of the morphisms $\emptyset$ since the morphisms come with a domain and codomain) This may seem strange, but the definition requires it since there are no $a \in \emptyset$ so the needed condition vacuously holds for $f = \emptyset$. In particular, in Set for every object $B$ there is a unique morphism $f:\emptyset \to B$ (In technical language, this says that $\emptyset$ is an initial object in Set). So, yes, in Set, $\emptyset$ has an identity morphism $\emptyset \to \emptyset$ and also participates in morphisms to all other sets. This means your intuition about the category Set is incorrect.
In general, a category is really just a collection of objects with arrows (morphisms) satisfying the right rules. Anything you write down with those properties is fine! We could consider the category whose objects are sets with just the identity arrows, or with all possible arrows; both would be fine. The advantage of this is that it lets lots of things fit into this framework.