Expressing that a functor is natural
Being a functor requires two properties:


fmap id = id
fmap (f . g) = fmap f . fmap g


The definition fmap fails the first condition:

fmap id = map id . reverse = reverse != id


So that's not a functor at all!

And it even fails the second condition:

fmap (f . g) = map (f . g) . reverse
             = map f . map g . reverse
             = map f . fmap g
             = reverse . map f . reverse . fmap g
             = reverse . fmap f . fmap g
             != fmap f . fmap g


Which is quite clear if you think about it:


fmap (f . g) reverses the list once and then maps its elements via f . g
fmap f . fmap g reverse the list, maps via g, then reverses the list again and then maps via f!