Isn't every Endofunctor an identity Functor? I mean Endofunctor maps category $A$ to category $A$ - so all of it's objects, and morphisms must remain the same. Isn't this an identity Functor?
For example, if I have a category $A$ of integers with an ordering on them:

$O = \{1,2,3,4\}$
  $M(x,y) = \{(x,y) |  x,y \in O, x \leq y)\}$

any Endofunctor I try to think of ends up being an identity Functor...
What I am thinking wrong here? Thanks.
EDIT: As per suggestion, let me try mapping all objects, via and endofunctor $F$ to $1$ and all morphisms to $1 \leq 1$.
$$F_{O}: O_{A} \to O_{A}, F_M: M_{A} \to M_{A}$$
then $F_{O}(1) = 1, F(2) = 1, ...$ and $F_{M}( M(1,2) ) = M(1,1) $
so now either:


*

*Nothing is changed and $A$ remains the same (endofunctor having only "mapped" things around)

*Or the new $A$ is now missing all objects except for $1$ and all morphisms except for $1\leq 1$.


If it's the former (1), what is the point of this functor in the first place, if it does  not "do" anything...
 A: No. The objects and morphisms don't have to remain the same.
An endofunctor is just a functor of the form $A\to A$, for a category $A$.
For example, $X\mapsto \{0,1\}\times X\ $  is an endofunctor on $Set$.
In your example also any constant mapping will give an endofunctor, also ($1\mapsto 1,\ 2\mapsto 1,\ 3\mapsto 4,\ 4\mapsto 4$) works...
A: You seem to give extra constraints to the definition of "endofunctor". In reality it is just a functor from/to the same category. It does not have to be an identity.In particular, note that an endofunctor does not have to be full or faithful or injective/surjective on objects.  Consider - as a parallel - a set $A$ and a function $f: A \to A $. Is $f$ necessarily the identity function? Of course not! It could be a permutation on set A or a constant function, or...
A: Take the category of "data structures", it's objects are basic types like numbers $\mathbb N$, booleans $\mathbb B$, pairs of objects $(a,b)$ and lists of objects $[a]$.
An example of a functor is the list functor: if we have a morphism like $\text{odd} : \mathbb N \longrightarrow \mathbb B$ then the list functor gives us $[\text{odd}] : [\mathbb N] \longrightarrow [\mathbb B]$ (which computes [F,T,F,T] from [0,1,2,3])
A: For the same reason, every "endomap" of sets $f: X \longrightarrow X$ would always be an identity, wouldn't it?
So, for instance, the only function $f: \mathbb{R} \longrightarrow \mathbb{R}$ would be the identity function, $f(x) = x$. And things like $f(x) = x^2, \sin x , e^x \dots$ and the like strange, inexplicable phenomena.  :-)
