That is, unpacking the definitions, are all morphisms $f : A \rightarrow B$ with retraction $r : B \rightarrow A$ the equalizer of some morphism $k : B \rightarrow C$ with the unique zero morphism $0_{BC} : B \rightarrow C$?
The ncatlab says yes [edit: apparently no it doesn't, as pointed out below], as they assert that the types of mono and epi form a total order with no qualifiers, but I haven't seen any proof of this – which is weird, because it seems an essential thing to prove (as proving that split monos are all normal would prove that they are also regular, strong, extremal, etc). Nor is a proof completely obvious to me – the proof that all split monos are regular involves proving that they equalize $f \circ r$ with the identity $1_B$, but both of these seem far from being zero morphisms.
I suspect that the solution might be that if $f \circ r$ is not the identity map in a category with zero morphisms, i.e. $f$ is not an isomorphism, then it necessarily follows that it must be a zero morphism itself, but I haven't had any luck proving that or coming up with counterexamples. It's clear to me that if $f$ is not an isomorphism, then $ f \circ r$ cannot be a monomorphism or an epimorphism either, which is suggestive.
Best Answer
It is not true. For example, in the category of groups normal monomorphisms are just injective morphisms with normal image. Now if $G$ is a nonabelian group, the diagonal $$\delta:G\to G\times G:g\mapsto (g,g)$$ is obviously a split monomorphism, but its image is not a normal subgroup of $G\times G$, so it is not a normal monomorphism.