Is the free cocompletion conservative? For a category $C$, $Set^{C^{op}}$, the category of presheaves on $C$, can be considered the free cocompletion of $C$. I wonder:

If $Set^{C^{op}}$ is equivalent to $Set^{D^{op}}$, does it follow that $C$ is equivalent to $D$?

 A: The phrase you're looking for is "Morita equivalence (of categories)" and the answer is no. All you can say is that the Cauchy completions (or more particularly, the Karoubi envelopes) of $\mathcal C$ and $\mathcal D$ are equivalent (see this Mathoverflow question).
The idea is that every presheaf preserves certain limits called absolute limits. This implies that adding these limits to the category (if they don't already exist) changes the category, but not the category of presheaves. Extending the presheaves to these new limits doesn't add any information because the presheaves have to preserve these new limits anyway.
The main example of an absolute limit is the splitting of an idempotent. This is a morphism from an object to itself satisfying $f \circ f = f$. That the idempotent splits means there are two further morphisms $r$ and $s$ (this time they don't necessary have the same domain and codomain) such that $s \circ r = f$ and $r \circ s = \operatorname{id}$.
It's easy to come up with a category where not every idempotent splits (consider the minimal category with an idempotent). Then adding in the splitting (the two extra morphisms) gives an inequivalent category whose presheaves are equivalent.

Conservativity of a functor $F \colon \mathcal C \to \mathcal D$ is the property that for any morphism $g$ in $\mathcal C$, if $F(g)$ is an isomorphism, then so is $g$. As Zhen Lin pointed out in the comments, this isn't quite the same as saying that if $F(c) \cong F(c')$, then $c \cong c'$ because the isomorphism $F(c) \cong F(c')$ doesn't necessarily come from a morphism in $\mathcal C$. For an example, consider the inclusion of the discrete two-object category into the category with two objects and an isomorphism between those objects.
Now presumably you want the "pseudo" version of conservativity so that isomorphisms are replaced with equivalences. Regarded as a pseudofunctor from small categories to locally small cocomplete categories, $\operatorname{PSh}$ takes a functor $F$ to the Yoneda extension of $Y \circ F \colon \mathcal C \to \operatorname{PSh}(\mathcal D)$. There are several ways to think of this more explicitly; the one I prefer is $\operatorname{PSh}(F)(G)(d) = \int^{c \in C} \mathcal D(d, F(c)) \times G(c)$, written using a coend. It can also be written just using colimits.
So the question becomes: if $\operatorname{PSh}(F)$ is an equivalence, is $F$ also an equivalence?
The same example shows that this is false too. There's an inclusion from the category without split idempotents to the category with them and this induces an equivalence between the two presheaf categories. But the inclusion is not itself an equivalence.
A: No, equivalency of the presheaf categories need not entail equivalency of the categories $C$ and $D$.
Fix a category $C$ and construct its idempotent completion, i.e., the universal functor $C\to S(C)$ to the category of categories where all idempotents split. Concretely, $S(C)$ has as objects pairs $(X,x)$ with $X$ an object in $C$ and $x\colon X\to X$ an endomorphism. The morphisms in $S(C)$ are $f\colon (X,x)\to (Y,y)$ where $f\colon X\to Y$ is a morphism in $C$ satisfying $f\circ x = f = y\circ f$. Composition is inherited from $C$ (but the identities are not). Let $i\colon C\to S(C)$ be given by $i(X)=(X, \mathrm {id}_X)$. Clearly, $C$ and $S(C)$ need not be equivalent.
It is not difficult to see (and best to argue by the property of idempotents being split) that if $E$ is any category in which idempotents split, then the functor $\mathbf E^{S(C)}\to \mathbf E^{C}$ is an equivalence of categories. In particular, the categories of presheaves on $C$ and on $S(C)$ are equivalent categories.
