Does Wolfram say all functional graphs have countably many vertices? Does Wolfram here say all functional graphs have countably many vertices?
Let a functional graph $G$ map the orbit of the following function $f:\Bbb Z_p\to\Bbb Z_p$
$x\mapsto p^{-1}\cdot(x-(x\pmod p))$ through the $p-$adic integers.
(Here $x\pmod p$ is the function that maps to the natural number $n$ representative of some element of $\Bbb Z/p\Bbb Z$ which is $0\leq n<p$)  So you can think of $f$ as simply deleting the most significant digit of any $p-$adic number.
Then this functional graph comprises uncountably many connected subgraphs.  The ones with eventually periodic orbits of period $m$ can be represented by the base $p$ Lyndon words of length $m$. Then there are uncountably many left over, which one needs the axiom of choice to pick representatives of.
I'm confused by where Wolfram says;

and can therefore be specified by a function mapping $\{1,...,n\}$ onto itself

which appears to imply a functional graph must be countable.  How does this sentence apply to my example? As I'm pretty sure I'm misunderstanding it.
 A: The functional graphs in the MathWorld article wouldn't just be countable; they'd be finite. This is not surprising to see in that article:

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*Many, perhaps most discussions of graph theory only consider finite graphs; many results about finite graphs do not extend to infinite graphs, or take considerably more work to do so. If I read anything about graph theory, I would not assume it applies to infinite graphs unless this is specifically mentioned.

*In particular, if 50% of the article is devoted to implementation details in Mathematica, which cannot possibly hope to deal with infinite graphs, I think it's safe to say that it only considers the finite case.

Nevertheless, there's nothing wrong with talking about an infinite functional graph; it can be specified by a function mapping the vertex set $V$ to itself, but there's nothing inherently wrong with $V$ being infinite. Some definitions for finite graphs break down for infinite graphs before you even try to prove anything about them, but this isn't one of them.
Your example is a perfectly good uncountably infinite functional graph, though there are two warnings that should always be given in this situation:

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*Be careful about applying any particular theorems about functional graphs you encounter, before you check if those theorems make sense in the infinite (and the uncountable) settings.

*If you're discussing it with graph theorists, you should be upfront about the vertex set being uncountably infinite if you want to avoid any confusion.

