Natural transformations of identity determined by endos of projective generator in abelian category? $\newcommand{\cat}[1][C]{\mathcal{#1}}$
Let $\cat$ be a finite abelian category, let $G\in \cat$ be a projective generator, and denote $E=\operatorname{End}(G)$.
I want to understand the (probably very easy) statement

$\eta\in \operatorname{End}(1_{\cat})$ is determined by $\eta_G \in Z(E)$.

So more precisely, I want to understand: how to define a natural transformation given $f\in Z(E)$?
Some very vague guess I have is using an equivalence $\cat \cong E\text{-mod}$, whence
\begin{align}
   \operatorname{End}(1_{\cat}) \cong
   \operatorname{End}(1_{E\text{-mod}}) \cong
   Z(E)
\end{align}
where the second step is easy.
But I don't understand the equivalence.
My question is then two-fold:


*

*Is the obvious functor $\cat(G,-): \cat \to E\text{-mod}$ an equivalence, and if yes, what is its inverse (as explicit as possible)?

*Given $f\in Z(\operatorname{End}(G))$, what does the corresponding natural endotransformation of $1_{\cat}$ look like?
Cheers

The statement is true in more generality: an element $\eta \in \text{End}(1_C)$ is determined by how it behaves on a generator $P$, in any category whatsoever, with no projectivity or linearity hypotheses. 
To see this, let $\eta, \eta' \in \text{End}(1_C)$ such that $\eta_P = \eta_P' : P \to P$. If $X$ is any other object and $f : P \to X$ any morphism, then by naturality we have
$$\eta_X \circ f = f \circ \eta_P = f \circ \eta_P' = \eta_X' \circ f.$$
Since $P$ is a generator, it follows that $\eta_X = \eta_X'$. $\Box$
This only says that the map $\text{End}(1_C) \to Z(\text{End}(P))$ is injective, though, not that it's surjective. 
$\text{End}(1_C)$ is sometimes called the center of $C$ and denoted $Z(C)$; see this blog post for more.
