If F is representable, is G?
Also no, because this converse is actually
equivalent to the original.
Since
H:C≃D then it has an inverse
H−1:D≃C.
Whiskering α by this inverse we get
αH−1:GHH−1≃FH−1
since
HH−1≅idD then it’s probably true that
α also acts as an
equivalence
αH−1:G≃FH−1
whose inverse is an
equivalence (αH−1)−1:FH−1≃G.
By this correspondence
H↦H−1;α↦(αH−1)−1 we see that
F and
G are symmetric in the problem statement and so having already answered part one of the question we have also answered part two.