Are there non-scalar endomorphisms of the functor $V\mapsto V^{**}/V$? Let $K$ be a field. Are there non-scalar endomorphisms of the endofunctor 
$$
V\mapsto V^{**}/V
$$ 
of the category of $K$-vector spaces? 
I asked a related question on Mathematics Stackexchange, but got no answer.
EDIT (Apr 15'14). Here is a closely related question which involves only basic linear algebra (and no category theory):
Let $K$ be a field and $V$ an infinite dimensional $K$-vector space. Is the $K$-algebra 
$$
\operatorname{End}_{\operatorname{End}_K(V)}(V^{**}/V)
$$
isomorphic to $K$?
EDIT (Apr 26'14). To avoid any misunderstanding, let me say explicitly that I'm unable to prove any of the following two statements:
(1) There is a pair $(K,V)$, where $K$ is a field and $V$ an infinite dimensional $K$-vector space, such that 
$$
\operatorname{End}_{\operatorname{End}_K(V)}(V^{**}/V)\simeq K.
$$
(2) There is a pair $(K,V)$, where $K$ is a field and $V$ an infinite dimensional $K$-vector space, such that 
$$
\operatorname{End}_{\operatorname{End}_K(V)}(V^{**}/V)\not\simeq K.
$$
EDIT (May 17'14). Here is a slight amplification of the above edit:
For any field $K$ and any infinite cardinal $\alpha$ put
$$
d(K,\alpha):=\dim_K\operatorname{End}_{\operatorname{End}_K(V)}(V^{**}/V),
$$
where $V$ is an $\alpha$-dimensional $K$-vector space.
Let $\kappa$ be the cardinal of $K$. The Erdős-Kaplansky Theorem implies
$$
1\le d(K,\alpha)\le\kappa\wedge(\kappa\wedge(\kappa\wedge\alpha)),
$$
where $\wedge$ denotes exponentiation.
There is no pair $(K,\alpha)$ for which I can prove that the first inequality is strict, and no pair $(K,\alpha)$ for which I can prove that the second inequality is strict.
EDIT (Jun 12'14). Here is a positive result. Unfortunately, it is very weak, and I hope users will be able to improve it.
Recall that $K$ is a field. For any vector space $V$ put $V':=V^{**}/V$. Let $V$ be an infinite dimensional vector space, and form the $K$-algebras
$$
A:=\operatorname{End}_KV,\quad B:=\operatorname{End}_KV',\quad C:=\operatorname{End}_AV',
$$
so that $C$ is the commutant of $A$ in $B$. Recall that the dimension of $B$ is
$$
\dim B=\operatorname{card}K\wedge\Big(\operatorname{card}K\wedge\big(\operatorname{card}K\wedge\dim V\big)\Big),
$$
where $\wedge$ denotes exponentiation. We claim

The codimension $c$ of $C$ in $B$ satisfies $\aleph_0\le c\le\dim B$.

To prove this, it suffices to show that, for each integer $n\ge2$, there is a subspace $W$ of $B$ satisfying $\dim W/(W\cap C)\ge n$.
Let $n$ be $\ge2$, let $V_1,\dots,V_n$ be subspaces of $V$ such that $V=V_1\oplus\cdots\oplus V_n$ and $\dim V_i=\dim V$ for all $i$, and let us define isomorphisms $\varphi_{ij}:V_j\to V_i$ for $1\le i,j\le n$ as follows.
Firstly $\varphi_{ii}$ is the identity of $V_i$. Secondly $\varphi_{i+1,i}:V_i\to V_{i+1}$, for $1\le i < n$, is any isomorphism. Thirdly $\varphi_{ij}$ for $j < i$ is the appropriate composition of the $\varphi_{k+1,k}$ previously defined. Fourthly we set $\varphi_{ij}:=(\varphi_{ji})^{-1}$ for $i < j$.
Define the morphism of $K$-algebras $\psi:M_n(K)\to B$ by
$$
\psi(a)(v)_i:=\sum_ja_{ij}\ \varphi'_{ij}(v_j)
$$
for $v=v_1+\cdots+v_n\in V'$ with $v_i\in V'_i$. Then define $W$ as the image of $\psi$. It is easy to see that $\psi$ is injective and that $W\cap C$ is the image under $\psi$ of the scalar matrices, so that $\dim W/(W\cap C)=n^2-1\ge n$.
There is not a single case in which I'm able to improve the above inequalities $\aleph_0\le c\le\dim B$.
 A: I've thought about this problem on and off for the last couple years.  While I haven't solved it, I did have one comment which is too big to fit in the comments section.
Let $K$ be a field and let $V_K$ be an infinite-dimensional $K$-vector space.  Let $A={\rm End}(V_K)$ as above.  We may identify $V_K$ with $K^{(I)}$ for some index set $I$; this is the set of columns vectors, with entries indexed by $I$, such that only finitely many entries are non-zero.  [If we think of $V=K[x]$, then this identifies $x^n$ with the $\omega$-indexed columns whose only non-zero entry is $1$ in the $n$th coordinate.]
The ring $A$ may then naturally be identified as the ring of $I\times I$ column-finite matrices.  This is the set of $I\times I$ matrices, such that each column has only finitely many non-zero entries.  The two-sided ideals in this ring form a well-ordered ascending chain, and the minimal non-zero ideal is the set of matrices of finite rank.  Call this ideal $J$.
The ideal $J$ is idempotent.  In fact, each element in $J$ is fixed, by left-multiplication, by a finite-rank idempotent (namely, the idempotent with $1$'s down the diagonal sufficiently far).  It is not difficult to prove that $JV^{\ast\ast}=V$.  Thus the left $A$-module structure on $V^{\ast\ast}/V$ has $J$ acting as zero.  Therefore, it really is an $A/J$-module structure.
When $V$ is countable dimensional, $J$ is also the unique maximal two-sided ideal (but there are many maximal one-sided ideals).  In particular, $A/J$ is simple in that case.  I don't know if this helps solve the problem in any way.
