It seems obvious to me that the set of functions with the signature $\forall A. A \rightarrow A$ is "once-inhabited", i.e. there is only one such polymorphic function which "works" for any set $A$, which is the identity function. (I think perhaps mathematicians would consider this to be actually a whole family of functions; I'm coming from CS where we tend to talk about single polymorphic functions, but feel free to explain how a mathematician would phrase the problem). However I don't know how to prove it except in a very hand-wavy manner.
A slightly more complex example would be to show that $$\forall A B C. (A \rightarrow B) \rightarrow (B \rightarrow C) \rightarrow (A \rightarrow C)$$ is also once-inhabited, and that the only such function with this signature is the function composition operator.
If you know about $\lambda$-calculus, you can make this precise by looking for normalized $\lambda$-terms that have type $\forall A. A \rightarrow A$ in the system F type system (which is described in details on https://fr.wikipedia.org/wiki/Système_F -- the English page is not as complete, for example it doesn't have the rules for typing).
In this basic case, looking at the inference rules, you quickly get that $\Lambda A. \lambda x:A. x$ is the only normalized term having the type $\forall A. A \rightarrow A$.
The Wikipedia page also explains how to make a boolean type (a type inhabited by exactly 2 terms), a natural number type, product types, and so forth.