(This question originates from Pinter's Book of Abstract Algebra Chapter 17 exercise F2.)
Let $G$ be an abelian group in additive notation. An $\textit{endomorphism}$ of $G$ is a homomorphism from $G$ to $G$. Let $\operatorname{End}(G)$ denote the set of all the endomorphisms of $G$, and define addition and multiplication of endomorphisms as follows: \begin{align*} [f+g](x) &= f(x) + g(x) & \text{for every }x\text{ in }G \\ fg &= f\circ g & \text{the composite of }f\text{ and }g \end{align*}
List the elements of $\operatorname{End}(\mathbb{Z}_4)$, then give the addition and multiplication tables for $\operatorname{End}(\mathbb{Z}_4)$.
My attempt:
$\quad\operatorname{End}(\mathbb{Z}_4) = \{\phi, \epsilon, \kappa, \tau\},\quad$ where
$\quad\phi = \left(\begin{matrix} 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{matrix}\right)\quad\epsilon = \left(\begin{matrix} 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 \end{matrix}\right)$
$\quad\kappa = \left(\begin{matrix} 0 & 1 & 2 & 3 \\ 0 & 3 & 2 & 1 \end{matrix}\right)\quad\tau = \left(\begin{matrix} 0 & 1 & 2 & 3 \\ 0 & 2 & 0 & 2 \end{matrix}\right)$
$\quad\begin{array}{l|c c c c} + & \phi & \epsilon & \kappa & \tau \\ \hline \phi & \phi & \epsilon & \kappa & \tau \\ \epsilon & \epsilon & \tau & \phi & \kappa \\ \kappa & \kappa & \phi & \tau & \epsilon \\ \tau & \tau & \kappa & \epsilon & \phi \\ \end{array}$ $\qquad\begin{array}{l|c c c c} \circ & \phi & \epsilon & \kappa & \tau \\ \hline \phi & \phi & \phi & \phi & \phi \\ \epsilon & \phi & \epsilon & \kappa & \tau \\ \kappa & \phi & \kappa & \epsilon & \tau \\ \tau & \phi & \tau & \tau & \phi \\ \end{array}$
Does this look right?
Everything you wrote is correct, but from the wording of the question (I do not have access to the book at the moment) it is unclear to me if you also need to prove that these are all the endomorphisms or not. In case you do, you can proceed as follows:
Since $(\mathbb{Z}_4,+) = \{ 0, x, 2x, 3x \}$ is cyclic, and $\phi(x+y) = \phi(x)+\phi(y)$ then every endomorphism is completely determined by what element $\phi(x)$ is.
In particular, you have up to four choices: $0, x, 2x, 3x$. Now you have exhibited a homomorphism for each of these choices, which in particular implies they are all possible.