What does it mean to compose two Functors?

Question

Exercise 5 of the Haskell Typeclassopedia Section 3.2 asks for a proof or counterexample on the statement

The composition of two Functors is also a Functor.

I thought at first that this was talking about composing the fmap methods defined by two separate instances of a Functor, but that doesn't really make sense, since the types wouldn't match up as far as I can tell. For two types f and f', the types of fmap would be fmap :: (a -> b) -> f a -> f b and fmap :: (a -> b) -> f' a -> f' b, and that doesn't really seem composable. So what does it mean to compose two Functors?

Answer 1

A Functor gives two mappings: one on the type level mapping types to types (this is the x in instance Functor x where), and one on the computation level mapping functions to functions (this is the x in fmap = x). You are thinking about composing the computation-level mapping, but should be thinking about composing the type-level mapping; e.g., given

newtype Compose f g x = Compose (f (g x))

can you write

instance (Functor f, Functor g) => Functor (Compose f g)

? If not, why not?