To (attempt to) answer the actual question (which is less interesting than the answers to the broader issues raised), the question is ill formed because of a "type error"
Either ~ (+)
(,) ~ (*)
(->) b ~ flip (^)
() ~ 1
Void ~ 0
These all map types to integers, and type constructors to functions on naturals. In a sense, you have a functor from the category of types to the category of naturals. In the other direction, you "forget" stuff, since the types preserve algebraic structure while the naturals throw it away. I.e. given Either () ()
you can get a unique natural, but given that natural, you can get many types.
But this is different:
(forall r . (a -> r) -> r) ~ a
It maps a type to another type! It is not part of the above functor. It's just an isomorphism within the category of types. So let's give that a different symbol, <=>
Now we have
(forall r . (a -> r) -> r) <=> a
Now you note that we can not only send types to nats and arrows to arrows, but also some isomorphisms to other isomorphisms:
(a, (b, c)) <=> ((a, b), c) ~ a * (b * c) = (a * b) * c
But something subtle is going on here. In a sense, the latter isomorphism on pairs is true because the algebraic identity is true. This is to say that the "isomorphism" in the latter simply means that the two types are equivalent under the image of our functor to the nats.
The former isomorphism we need to prove directly, which is where we start to get to the underlying question -- is given our functor to the nats, what does forall r.
map to? But the answer is that forall r.
is neither a type, nor a meaningful arrow between types.
By introducing forall, we have moved away from first order types. There's no reason to expect that forall should fit in our above Functor, and indeed, it doesn't.
So we can explore, as others have above, why the isomorphism holds (which is itself very interesting) -- but in doing so we've abandoned the algebraic core of the question. A question which can be answered, I think, is, given the category of higher-order types and constructors as arrows between them, what is there meaningful Functor to?
Edit:
So now I have another approach which shows why adding polymorphism makes things go nuts. We start by asking a simpler question -- does a given polymorphic type have zero or more than zero inhabitants? This is the type inhabitation problem, and winds up being, via Curry-Howard, a problem in modified realizability, since it's the same thing as asking if a formula in some logic is realizable in an appropriate computational model. Now as that page explains, this is decidable in the simply typed lambda calculus but is PSPACE-complete. But once we move to anything more complicated, by adding polymorphism for example and going to System F, then it goes to undecidable!
So, if we can't decide if an arbitrary type is inhabited at all, then we clearly can't decide how many inhabitants it has!