TITLE: Lax Functors and Equivalence of Bicategories?
QUESTION [21 upvotes]: Lax functors of bicategories were introduced at the very inception of bicategories, and I'm trying to get a better feel for them. They are the same as ordinary 2-functors, but you only require the existence of a coherence morphism, not an isomorphism. The basic example I'm looking at are when you have a lax functor from the singleton bicategory to a bicategory B. These are just object b in B with a monad T on B. 
My Question: If I have an equivalence of bicategories A ~ A', do I a get equivalent bicategories of lax functors Fun(A, B) and Fun(A', B)? If not, is there any relation between these two categories? 
[Edit] 
So let me be more precise on the terminology I'm using. I want to look at lax functors from A to B. These are more general then strong/pseudo and much more general then just strict functors. For a lax functor we just have a map like this:
$ F(x) F(g) \to F(fg)$
for a strong or pseudo functor this map is an isomorphism, and for strict functor it is an identity. I don't care about strict functors. 
I'm guessing that these form a bicategory Fun(A,B), with the 1-morphism being some sort of lax natural transformation, etc, but I don't really know about this. Are there several reasonable possibilities?
When I said equivalence between A and A' what I meant was I had a strong functor F:A --> A' and a strong functor G the other way, and then equivalences (not isomorphisms) FG = 1, and GF = 1. This seems like the most reasonable weak notion of equivalence to me, but maybe I am naive.
I haven't thought about equivalences using lax functors. Would they automatically be strong? What I really want to understand is what sort of functoriality the lax functor bicategories Fun(A, B) have?

REPLY [10 votes]: I was surprised (and delighted) to discover that the answer is no.  Here is an example; in fact it's a special case of the example Mike referred to, but worked out in more detail.
Let S be a set and cd(S) the codiscrete category on S (Hom(x, y) = • for every x and y in S).  Let BSet denote the one-object bicategory corresponding to the monoidal category (Set, ×).  As remarked here on the nlab, a lax functor cd(S) → BSet is the same data as a category with object set S.  However, there is a catch: the morphisms (whether lax or strong natural transformations) are not just functors between categories fixing the objects.  Instead, if C and D are categories corresponding to two lax functors cd(S) → BSet, then a morphism (X, F) from C to D consists of

for each object s of S, a set X(s),
for each pair of objects s and t of S, a map F(s, t) : C(s, t) × X(t) → X(s) × D(s, t),
such that obvious identity and composition laws hold.

Depending on which notion of natural transformation you want to take, the map F(s, t) might be an isomorphism, or it might go the other way.  I'll consider all three possibilities simultaneously.
We can compose 1-morphisms: the composition (Z, H) of (X, F) and (Y, G) has Z(s) = X(s) × Y(s), and H formed from F and G in an obvious way.  The identity morphism has X(s) = • and F(s, t) = id.
We also have 2-morphisms from (X, F) to (Y, G), which are families of maps X(s) → Y(s) making some diagrams involving X, F, Y, G commute.
This doesn't yet seem to be any familiar 2-category.  Let's extract the maximal 2-groupoid by taking only invertible 1- and 2-morphisms.  (That's certainly an equivalence-invariant thing to do.)  The invertible 2-morphisms are the ones where the maps X(s) → Y(s) are isomorphisms.  For (X, F) to be invertible, we first need for every s, an object Y(s) and an isomorphism X(s) × Y(s) = •.  That can only happen if X(s) = • for every s.  (This is the only property of (Set, ×) I really care about: it has no invertible objects besides •.)  Now F is just an ordinary functor C → D which is the identity on objects, and for (X, F) to be invertible F needs to be an isomorphism.  When (X, F) and (Y, G) are invertible, so X(s) = Y(s) = •, there is at most one 2-morphism between them, exactly when F and G are equal as maps C(s, t) → D(s, t).  In conclusion, the maximal 2-groupoid contained in the 2-category of lax functors cd(S) → BSet is the 1-groupoid of categories with object set S and isomorphisms fixing the objects.  Note that it didn't matter for the argument which notion of natural transformation I chose.
Now as S varies over nonempty sets, these resulting groupoids are nonequivalent categories.  However, surely the 2-categories cd(S) are all equivalent under any reasonable definition, since they equivalent 1-categories.