You're right, those two functors \$$F\$$ and \$$F'\$$ are not isomorphisms, but they are *related* by an isomorphism: \$$F' = F \circ S\$$, where \$$S: \mathbf{M}\to\mathbf{M}\$$ is the functor represented by the sequence \$$(3, 2, 5, 7, 11, 13, \dots)\$$, which is its own inverse. What I think is true is that two functors \$$F_1, F_2\$$ are "equivalent" in the sense you describe, having the same image, just reordered, exactly when \$$F_1=F_2\circ S\$$ for some invertible functor \$$S\$$.