Even the real progenitors of a lot of this sort of thought, like E.T. Jaynes, expoused significantly more skepticism than I've ever seen a "rationalist" use. I would even imagine if you asked almost all rationalists who E.T. Jaynes was (if they weren't well versed in statistical mechanics) they'd have no idea who he was or why his work was important to applying "Bayesianism".
It would surprise me if most rationalists didn't know who Jaynes was. I first heard of him via rationalists. The Sequences talk about him in adulatory tones. I think Yudkowsky would acknowledge him as one of his greatest influences.
Obviously, since f(f) = f it should be a -> a as well. But to infer it without actually evaluating it, using the standard Hindley-Milner algorithm, we reason as follows: the two f's can have different specific types, so we introduce a new type variable b. The type of the first f will be a substitution instance of a -> a, the type of the second f will be a substitution instance of b -> b. We introduce a new type variable c for the return type, and solve the equation (a -> a) = ((b -> b) -> c), using unification. This gives us the substitution {a = (b -> b), c = (b -> b)}, and so we see that the return type c is b -> b.
But if we use pattern matching rather than unification, the variables in one of the two sides of the equation (a -> a) = ((b -> b) -> c) are effectively treated as constants referring to atomic types, not variables. Now if we treat the variables on the left side as constants, i.e. we treat a as a constant, we have to match b -> b to the constant a, which is impossible; the type a is atomic, the type b -> b isn't. If we treat the variables on the right side as constants, i.e. we treat b and c as constants, then we have to match a to both b -> b and c, and this means our substitution will have to make b -> b and c equal, which is impossible given that c is an atomic type and b -> b isn't.