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 $id_A = 1_A$.
Your statements 1 and 3 are correct, but note that you don't even need injectivity on objects to prove that your functor is faithful.
Your statement 2 is incorrect, however. For example, you could take $\mathcal{C}$ and $\mathcal{D}$ to be discrete categories, then if $X\neq Y$ are objects in $\mathcal{C}$ such that $F(X)=F(Y)$, you will have a morphism in $\mathcal{D}(F(X),F(Y))$ but none in $\mathcal{C}(X,Y)$, so in that case $F$ is not full.
In fact if you see your thin categories as preorders $(C,\leq_C )$ and $(D,\leq_D)$, then a functor is the same thing as a preorder-preserving map, i.e. a map such that
$$x\leq_C y\Rightarrow f(x)\leq_D f(y)$$and a full functor is the same thing as a pre-order-preserving and preorder-reflecting map, i.e. a map $f$ such that
$$x\leq_C y\Leftrightarrow f(x)\leq_D f(y).$$
Best Answer
If $X$ is an object in a category $\mathcal C$, the identity morphism $id_X:X\to X$ is a morphism in the category $\mathcal C$.
On the other hand, the identity functor is not a morphism in $\mathcal C$; it is a functor $\mathcal C\to\mathcal C$. A functor takes has two inputs, (i) objects, and (ii) morphisms. So the identity functor $Id_{\mathcal C}$ has, for every object $X$ in $\mathcal C$ and morphism $f:X\to Y$, the values $Id_{\mathcal C}(X)=X$ and $Id_{\mathcal C}(f)=f$.
I think it will be easier to see what's going on if you look at a specific category. So let's say $\mathcal C$ is the category of Sets. Then an object in $\mathcal C$ is just a set $X$, and a morphism $f:X\to Y$ is just a set function.
In this case, the identity morphism of a fixed set $X$ is just the identity map on $X$ that you're surely familiar with. It just has the value $id_X(x)=x$ for all $x\in X$.
On the other hand, a functor $F:\mathcal C\to\mathcal C$ assigns to every set $X$ a new set $F(X)$ and to every morphism $f:X\to Y$ a new morphism $F(f):F(X)\to F(Y)$. In the case of the identity functor, it just takes $X\mapsto X$ and $f\mapsto f$.