First of all, for any two bicategories A and B, there is a bicategory $Fun_{x,y}(A,B)$ where x can denote either strong, lax, or oplax functors, and y can denote either strong, lax, or oplax transformations. There's no problem defining and composing lax and oplax transformations between lax or oplax functors, and the lax/oplax-ness doesn't even have to match up. It's also true that two x-functors are equivalent in one of these bicategories iff they're equivalent in any other one. That is, any lax or oplax transformation that is an equivalence is actually strong/pseudo.
Where you run into problems is when you try to compose the functors. You can compose two x-functors and get another x-functor, but in general you can't whisker a y-transformation with an x-functor unless x = strong, no matter what y is, and moreover if y isn't strong, then the interchange law fails. Thus you only get a tricategory with homs $Fun_{x,y}(A,B)$ if x=y=strong. (In particular, I think this means that there isn't a good notion of "equivalence of bicategories" involving lax functors.)
For a fixed strong functor $Fcolon Ato A'$, you can compose and whisker with it to get a functor $Fun_{x,y}(A',B) to Fun_{x,y}(A,B)$ for any x and y. However, the same is not true for transformations $Fto F'$, and the answer to your question is (perhaps surprisingly) no! The two bicategories are not equivalent.
Consider, for instance, A the terminal bicategory (one object, one 1-morphism, one 2-morphism) and A' the free-living isomorphism, considered as a bicategory with only identity 2-cells. The obvious functor $A' to A$ is an equivalence. However, a lax functor from A to B is a monad in B, and a lax functor from A' to B consists of two monads and a pair of suitably related "bimodules". If some lax functor out of A' is equivalent to one induced by composition from A (remember that "equivalence" doesn't depend on the type of transfomation), then in particular the two monads would be equivalent in B, and hence so would their underlying objects. But any adjunction in B whose unit is an isomorphism gives rise to a lax functor out of A', if we take the monads to be identity 1-morphisms, the bimodules to be the left and right adjoint, and the bimodule structure maps to be the counit and the inverse of the unit. And of course can have adjunctions between inequivalent objects.
By the way, I think your meaning of "equivalence" for bicategories is becoming more standard. In traditional literature this sort of equivalence was called a "biequivalence," because for strict 2-categories there are stricter sorts of equivalence, where you require either the functors to be strict, or the two composites to be isomorphic to identities rather than merely equivalent to them, or both. These stricter notions don't really make much sense for bicategories, though. For instance, in a general bicategory, even identity 1-morphisms are not isomorphisms, so if "equivalence" were to demand that FG be isomorphic to the identity, a general bicategory wouldn't even be equivalent to itself!
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