Considering categories C and D, and functor $F:C\to D$, and the identity functor $Id_C : C\to C$. Is there always a natural transformation $n: Id_C \to F$, where n(X) is just F(X)?
$$\require{AMScd} \begin{CD} X @>{f}>> Y\\ @Vn(X)VV @Vn(Y)VV \\ F(X) @>{F(f)}>> F(Y). \end{CD}$$
It seems to me that you should always be able to define said natural transformation and in this sense every Functor is/has a natural transformation from the identity functor of its domain category. But I haven't seen this confirmed anywhere and in fact have seen things that seem to suggest it is not necessarily the case. Am I missing something?
Inspired by this question. Functors that has a natural transformation from identity
EDIT: Thanks to Dan Doel for an solid example. I also just realized that the first comment was correct $F: C \to D$ and $Id_c: C \to C$ only have the same target if F is an endofunctors, and if not then the definition of a natural transformation isn't even well formed (which was probably the root of my confusion).
The answer to the question in the title is "no." For instance, on the category of sets, there is a functor:
$$ \begin{align} &F : \mathsf{Sets} → \mathsf{Sets} \\ &F_0(X) = Ø \\ &F_1(f) = id_Ø \end{align} $$
This is just a constant functor. Now, a natural transformation $\mathsf{Id} → F$ would require there to be functions from every set into the empty set. However, the only such function is the one from the empty set to itself.
This example can be generalized by using initial objects in other categories. Sometimes there will be a natural transformation $\mathsf{Id} → F$ in such scenarios. But if the initial object is strict, then the only way this can happen is if every object is initial, which would mean that the category is equivalent to the terminal category.