Let $G$ be a group. If there is a homomorphism $f:G\to G$ (special case of the codomain being arbitrary group), then the kernel $f^{-1}(id)$ is a normal subgroup of $G$.
But now the other way around: Start out with the existence of a normal subgroup $H$ of $G$. Is there necessarily a homomorphism $f:G\to G$ such that the kernel of $f$ is $H$?
That's false.
If $G=\mathbb{Z}$ then any homomorphism $f:\mathbb{Z}\rightarrow\mathbb{Z}$ takes the form $f(a)=ma$ for some $m\in\mathbb{Z}$. Clearly the kernel is trivial (unless $m=0$ then the kernel is everything). However for all $n\in\mathbb{N}$ we have that $n\mathbb{Z}$ is a normal subgroup of $\mathbb{Z}$. In particular $2\mathbb{Z}$ is not a kernel of any homomorphim from $\mathbb{Z}$ to itself.
However, it is possible to "correct" this statement. If you only request an homomorphism from $G$ to some other group. Given any normal subgroup $H$, the quotient homomorphism $f:G\rightarrow G/H$ which sends $g\mapsto g+H$ has $H$ as it's kernel. In other words, every normal subgroup is a kernel of some homomorphism, not necessarily from $G$ to itself.