Do normal group endomorphisms form a normal submonoid? 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.
 A: 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.
