Let's start with End$(\mathbb{Z})$: Suppose $f\colon \mathbb{Z}\to \mathbb{Z}$ is an endomorphism. Suppose I know the value of $f(1)$. Then $$f(2)=f(1+1)=f(1)+f(1),$$
so I know the value of $f(2)$. Similarly $$f(-1)=-f(1),$$ so I know the value of $f(-1)$ too.
More generally, for $n>0$ $$f(n)=f(1+1+1+\cdots+1)=f(1)+f(1)+f(1)+\cdots +f(1)=nf(1),$$
and $$f(-n)=-f(n)=-nf(1).$$
Putting all this together, we see that applying $f$ to some integer $n$ is just multiplying by $f(1)$. So every element of End$(\mathbb{Z})$ is just multiplication by some fixed integer. For $a\in \mathbb{Z}$, let us use the notation $[a]$ to denote multiplication by $a$. Thus End$(\mathbb{Z})$ is just the set of such $[a]$.
Let us see what arithmetic on End$(\mathbb{Z})$ looks like. First addition:
$$([a]+[b])n=[a]n+[b]n=an+bn=(a+b)n=[a+b]n,$$
for all integers $n$.
The first equality here is just the definition of "pointwise" addition. Namely to evaluate the sum of two endomorphisms on some integer $n$, we just evaluate them each on $n$ and then add.
The second equality is unpacking our notation: $[a]$ means multiplication by $a$. The third equality is just bracketing, and the last one is again the definition of $[a+b]$.
So addition on End$(\mathbb{Z})$ is the same as addition on the integers:$$[a]+[b]=[a+b]$$
Now consider multiplication:$$([a][b])n=[a]([b]n)=[a](bn)=abn=[ab]n.$$
The first equality here says that the composition of the function $[a]$ with the function $[b]$, applied to $n$, is simply what you get if you first apply $[b]$ to $n$, then apply $[a]$ to the result. In other words: Do $[b]$ then do $[a]$.
(Note: This may seem a bit backwards, but it is just how composition is defined, so that the argument can go on the right.)
The remaining equalities follow from the definitions of $[b],[a], [ab]$.
Thus we have $$[a][b]=[ab].$$
That is multiplication in End$(\mathbb{Z})$ is just ordinary multiplication of the corresponding integers.
Thus End$(\mathbb{Z})$ can be identified with $\mathbb{Z}$, under the correspondence $$[a]\leftrightarrow a.$$ Further addition and multiplication are the same in End$(\mathbb{Z})$ and $\mathbb{Z}$. We write End$(\mathbb{Z})\cong\mathbb{Z}$.
The same approach can calculate End$(\mathbb{Z}/6\mathbb{Z})$. Again the endomorphisms are multiplication by integers, and pointwise addition corresponds to ordinary addition, whilst composition corresponds to ordinary multiplication. The only difference this time is that multiplication by $a$ and multiplication by $a+6$ are the same: $$[a]=[a+6]$$ Thus we have End$(\mathbb{Z}/6\mathbb{Z})\cong\mathbb{Z}/6\mathbb{Z}$.
For a third example let us calculate End$(\mathbb{Z}^3)$. We can write elements of $\mathbb{Z}^3$ as column vectors: $$\left(\begin{array}{c}u\\v\\w\end{array}\right)$$
Given an endomorphism $f\colon \mathbb{Z}^3 \to \mathbb{Z}^3$ we have
\begin{eqnarray*}f\left(\begin{array}{c}u\\v\\w\end{array}\right)&=&f\left(u\left(\begin{array}{c}1\\0\\0\end{array}\right)+v\left(\begin{array}{c}0\\1\\0\end{array}\right)+w\left(\begin{array}{c}0\\0\\1\end{array}\right)\right)\\\\&=&
uf\left(\begin{array}{c}1\\0\\0\end{array}\right)+vf\left(\begin{array}{c}0\\1\\0\end{array}\right)+wf\left(\begin{array}{c}0\\0\\1\end{array}\right)\\\\&=&
u\left(\begin{array}{c}a\\b\\c\end{array}\right)+v\left(\begin{array}{c}d\\e\\f\end{array}\right)+w\left(\begin{array}{c}g\\h\\i\end{array}\right)\\\\&=&
\left(\begin{array}{ccc}
a&d&g\\
b&e&h\\
c&f&i
\end{array}\right)
\left(\begin{array}{c}u\\v\\w\end{array}\right)
\end{eqnarray*}
Here $$
\left(\begin{array}{c}a\\b\\c\end{array}\right)=f\left(\begin{array}{c}1\\0\\0\end{array}\right),\qquad
\left(\begin{array}{c}d\\e\\f\end{array}\right)=f\left(\begin{array}{c}0\\1\\0\end{array}\right),\qquad
\left(\begin{array}{c}g\\h\\i\end{array}\right)=f\left(\begin{array}{c}0\\0\\1\end{array}\right)$$
So we see that the endomorphisms of $\mathbb{Z}^3$ are precisely multiplication by $3\times 3$ matrices with integer entries. For a $3\times 3$ matrix $M$, let $[M]$ denote the endomorphism of $\mathbb{Z}^3$ given by multiplication by $M$.
What about addition? We have $$([M]+[N])v=[M]v+[N]v=Mv+Nv=(M+N)v=[M+N]v.$$
That is pointwise addition of endomorphisms corresponds to addtion of matrices.
For multiplication we have $$([M][N])v=[M]([N]v)=[M](Nv)=M(Nv)=(MN)v=[MN]v.$$
That is composition of endomorphisms corresponds to multiplication of matrices.
In summary End$(\mathbb{Z}^3)=M_3(\mathbb{Z})$.
