Terms of an algebraic theory that act as homomorphisms Say you have some kind of "algebraic" category $A$ with a forgetful functor $U : A \to \mathbf{Set}$ which has a left adjoint $F : \mathbf{Set} \to A$. The natural transformations $U \to U$ can be interpreted as the terms of the algebraic theory. For example, if $A=\mathbf{CRing}$, these are just polynomials. Suppose we consider the map $\Phi : \mathrm{Nat}(id_A,id_A) \to \mathrm{Nat}(U,U)$ that simply takes a natural transformation $\eta : id_A \to id_A$ and performs horizontal composition with the identity natural transformation $U \to U$. What is the interpretation of the image of $\Phi$? That is, what are the terms that actually act as morphisms of $A$?
Why do $\mathbf{Ab}$ and related categories enjoy the property that $\Phi$ is surjective, i.e. every term acts as a morphism?
 A: An algebraic theory in which every term is a homomorphism is called commutative: ncatlab.org/nlab/show/commutative+algebraic+theory.  I'm not sure that's quite what you were asking, because here "term" means "term in any number of variables", that is, natural transformation $U^n \to U$ for any $n$ (not just $1$).  For instance, in the theory of abelian groups, $+: A \times A \to A$ is a homomorphism for any abelian group $A$.  But commutative theories are certainly a useful class to consider, and a lot has been written about them. 
A: Given two adjunctions  $<F, G>, <F', G'>: \mathscr{B}\to \mathscr{A}$ from standard properties  follow the natural isomorphism $Nat(F, F') \cong  Nat(1, G\circ F') \cong  Nat(G', G)$.
In our case follow that there is a isomorphism between the monoid of the natural transformations  $U \Rightarrow U$ and the dual of the monoid of natural transformations  $L \Rightarrow  L$.
Let $\alpha : L \Rightarrow L$, given a set $S$ we have the commutative  diagram
$\begin{array}{ccc}
L(1) & \xrightarrow{\alpha_1 } & L(1) \\
\widehat{x} \downarrow    & &\downarrow  \widehat{x} \\
L(S)& \xrightarrow{\alpha_S } & L(S)
\end{array}$
where $x\in X$ is identified by the unique morphism $\widehat{x}: L(1) \to L(S)$ such that $\widehat{x}(1)=x$.
I claim that $\alpha \mapsto \alpha_1$ is a bijection between the natural transformations $\alpha: L \Rightarrow L$ and the morphisms $a: L(1) \to L(1)$ (i.e. the set $L(1)$),
The set of morphisms $\widehat{x}: L(1) \to L(S) $ is a epimorphic family, then from the diagram this map is injective,  given $a: L(1) \to L(1)$, from the  diagram define $\alpha_S: L(S) \to L(S)$ as the (unique) morphisms such that $\alpha_S(x)=\widehat{x}\circ a$.
