They are not equal. Part of the definition of equality of functions should be that they have the same domain and codomain. That is, functions $f : X \to Y$ and $g : Z \to W$ are equal iff $X = Z, Y = W$, and $f(x) = g(x)$ for all $x \in X = Z$.
As discussed in the link, this is not the definition of equality you get if you think of functions $f : X \to Y$ in terms of their graphs $\Gamma_f \subset X \times Y$, since any two empty functions have the same graph (the empty subset of the empty set). But you don't have to do that, and you shouldn't. From a category-theoretic point of view it simply does not make sense to ask whether two morphisms in a category are equal unless they have the same source and target. And when we pass from the graph of a function to the function itself, it's not enough just to know the graph as a subset of the set $X \times Y$; we also have to know what the two coordinate projections $\pi_X : X \times Y \to X$ and $\pi_Y : X \times Y \to Y$ are, in order to know what values the function takes. This is extra data, and for empty functions with different codomains this extra data is different. (Of course for empty functions $\pi_Y : \emptyset \times Y = \emptyset \to Y$ is just the original function again, so this is a bit circular. But this at least shows that the graph argument is not decisive.)
Here is an additional consideration which may be persuasive. An isomorphism between two morphisms $f : X \to Y$ and $g : Z \to W$ (in the arrow category of some category) is a pair of isomorphisms $h_1 : X \cong Z$ and $h_2 : Y \cong W$ such that $h_2 \circ f = g \circ h_1$. Whatever equality of functions ought to mean it should surely be the case that functions that are equal are isomorphic. But two empty functions with non-isomorphic domains are not isomorphic.
