Okay, based on the extensive discussion in comments, it seems that you want to consider the following:
Suppose that we have a well-defined collection of sets, which we want to make into a category by letting them be the objects, taking the collection of morphisms to be all set-theoretic functions between the two sets, and using regular composition and domain/codomain identifications. Can we prove that under these circumstances, for us to have a category then the categorical identity arrow must be the identity function for the set?
The key is that you have enough functions to "separate points". Given any $a,b\in A$, $a\neq b$, there exists a function $g\colon A\to A$ such that $g(a)\neq g(b)$. For example, define $g$ to be the function that maps $b$ to $a$, and maps everything else to $b$. (Compare this with the example I gave in the comments, where this does not hold).
So, fix a set $A$, and suppose that $f\colon A\to A$ is the arrow that satisfies the identity conditions (for all objects $B$ and $C$, and all arrows $g\colon A\to B$ and $h\colon C\to A$, $gf = g$ and $fh = h$).
Pick any $a$ and $b$ in $A$, $a\neq b$. Let $g$ be a function with $g(a)\neq g(b)$; then $gf(a) = g(a)$, so it follows that $f(a)\neq b$. This holds for every $b\in A-\{a\}$, so the only possibility is that $f(a)=a$. This holds for all $a\in A$, so $f$ must be the identity map.
You can generalize this to any set-based category in which you can either separate points, or "hit" any point: if for every object $A$ and every elements $a,b\in A$ with $a\neq b$, there either exists an object $B$ and a morphism $h\in\mathcal{C}(A,B)$ such that $h(a)\neq h(b)$; or else there exists an object $C$, and a morphism $g\in\mathcal{C}(C,A)$ for which there exists an element $c\in C$ such that $g(c)=a$; then the identity morphism of $A$ must be the identity map of $A$.
Indeed, suppose that $f$ is the identity morphism, and let $a\in A$. For each $b\in A$, $b\neq a$, either we have $B$ and $h$ as above, and $hf(a) = h(a)$ implies that $f(a)\neq b$; or else there exists $C$, $c$ and $g$ as above with $g(c) = a$. Then $fg = g$ gives that $f(a)\neq b$ (since $f(g(c))=b$ and $g(c)=a$ implies $a=b$). Either way, you get that for all $b\in A$ with $b\neq a$, $f(a)\neq b$. So the only possibility left is that $f(a)=a$. This holds for all $a\in A$, so $f=1_A$.
Note. In a sense, the condition is both necessary and sufficient, though for silly reasons: if the condition is not met by $A$ and $a$, then the identity map of $A$ cannot be the identity morphism, simply because the identity map of $A$ satisfies the given conditions: for all $b\neq a$ you have $1_A(a)\neq 1_A(b)$, and $1_A(a)=a$.
Added. This argument applies to categories such as topological spaces (because you always have the map from the $1$-element topological space to your toplogical space mapping the unique point to $a$); pointed topological spaces (the discrete 2-element pointed topological space maps the non-distinguished point to your favorite point); groups (you have maps from the cyclic group to any group, mapping the generator to your element $a$); and others. It's hard to make it work with posets as categories, because posets as categories are not really set-based categories (the objects are not usually sets and arrows set-theoretic functions between them); you can model them as set-based categories, but then the result need not hold: the example I gave in the comments can be thought of as the totally ordered set with two elements, for example, and here you don't have $id_A = 1_A$.
You're right, $F$ really is two functions. If you were being very formal, you might say a functor $F : \mathcal{C} \to \mathcal{D}$ is a pair $F=(F_0, F_1)$, where $F_0 : \mathrm{ob}(\mathcal{C}) \to \mathrm{ob}(\mathcal{D})$ and $F_1 : \mathrm{mor}(\mathcal{C}) \to \mathrm{mor}(\mathcal{D})$ are functions satsifying
- If $f : X \to Y$ in $\mathcal{C}$, then $F_1(f) : F_0(X) \to F_0(Y)$ in $\mathcal{D}$;
- $F_1(\mathrm{id}_X) = \mathrm{id}_{F_0(X)}$ for all $X \in \mathrm{ob}(\mathcal{C})$; and
- $F_1(g \circ f) = F_1(g) \circ F_1(f)$ for all $X \xrightarrow{f} Y \xrightarrow{g} Z$ in $\mathcal{C}$.
There are various encoding tricks you could use to completely remove ambiguity, but in day-to-day life there is no problem with just writing $F$ to denote both $F_0$ and $F_1$. But when using a proof assistant, say, the distinction has to be made.
Best Answer
The morphisms of a general category need not be functions in any sense.
We can regard a category as a (possibly huge) directed graph, equipped with a(n associative) composition operation on the edges.
If we have a partial order relation $R$ on a set $X$, then taking elements of $X$ for objects and taking exactly one arrow (morphism) $x\to y$ if $x\,R\,y$ defines a category, in which the single arrow $x\to x$ is the identity morphism of object $x$.
Note also that ${\rm id}_x$ needs to satisfy ${\rm id}_x\circ f=f$ and $g\circ {\rm id}_x=g$ for all morphisms $f$ ending at $x$ and for all $g$ starting at $x$.