What it says on the tin.
A group endomorphism $v\colon G\to G$ is called normal if $v(aba^{-1})=av(b)a^{-1}$ for all $a,b\in G$. Equivalently, the map $g\mapsto v(g^{-1})g$ is a group homomorphism. Equivalently, the image of this map commutes with the image of $v$.
$\operatorname{End}(G)$, all endomorphisms of G, is a monoid under composition. Let $M$ be the set of all normal endomorphisms of $G$. This forms a submonoid. It is normal if $vM=Mv$ for all $v\in\operatorname{End}(G)$. So is it normal?
If we restrict to automorphisms, the normal automorphisms form what are also known as the central automorphism group. This is known to be a normal subgroup, and to my understanding the proof relies on the center being characteristic. This doesn't generalize to endomorphisms, though.
No, it's not.
First the abstract reason, followed by an example.
First, $v\colon G\to G$ is normal if and only if $Sv*id$ is a normal endomorphism, where $S$ is the inversion map. This is in turn equivalent to the images of $v$ and $Sv*id$ commuting.
Suppose $G$ is indecomposable. Then $id=v*(Sv*id)$ implies that one of the two morphisms is a central automorphism. Since their images commute, the other is thus central.
Let $w\colon G\to Z(G)$. This gives a normal endomorphism. Then for any endomorphism $v$ the composition $wv$ is also a normal endomorphism with central image. So we want to know if there is a normal endomorphism $w'$ (depending on $v,w$) such that $vw'=wv$. If we can find $v,w$ with no such $w'$, then we have the claim
The obstruction is intuitively clear, though slightly bothersome to write out.
More specifically, if we had a $v$ such that $v(Z(G))\setminus Z(G) = v(Z(G))\setminus 1$; and a non-trivial central homomorphism $w$ with $wv$ nontrivial, then no such $w'$ can exist: necessarily $vw'$ is not central.
Using GAP I found the following example, though it may not be of minimal order.
$G=\textrm{SmallGroup}(32,11)=(C_4\times C_4)\rtimes C_2$. As a permutation group $G$ can be generated by $x=(1,5)(2,6)(3,4)(7,8)$ and $y=(1,8,7,3,4,2,6,5)$. The center is generated by $(1,6,4,7)(2,3,8,5)$.
Define $w$ by $w(x)=1$, $w(y)= (1,6,4,7)(2,3,8,5)$.
Define $v$ by $v(x)=1$, $v(y)=(2,3,8,5)$.
Then $w,v$ have the desired properties, and we conclude that the normal endomorphisms of $G$ are not a normal sub-monoid.