For an elliptic curve $E/k$, let $\alpha$ be any endomorphism over $\bar{k}$ in End($E$) and let $[n]$ be the multiplication-by-$n$ endomorphism. Show that $[n] \bullet \alpha=\alpha \bullet [n]$, i.e., $[n]$ is commutative with all endomorphisms.
So we have that $[n] : E \rightarrow E$ is given by $P \rightarrow nP$ is an endomorphism defined over the given space but what's the easiest way to show that $[n]$ is commutative with all endomorphisms?
By definition, an endomorphism of a complex elliptic curve $E$ is a holomorphic map $E\rightarrow E$, which fixes the origin. It turns out that this condition is enough to force it to be a homomorphism of abelian groups in the usual sense, i.e., it is a $\mathbb{Z}$-module homomorphism.
This definition makes sense for every algebraically closed field $k$, replacing "holomorphic" by "algebraic". Then we have
Theorem: Let $\phi\colon E(k)\rightarrow E(k)$ be an algebraic map, i.e., given by rational functions. Then $\phi$ is a group endomorphism if and only if $\phi$ fixes the origin.