Can 2 different coverages *on the same category* yield the same sheaf topos? It is possible to have distinct sites $(\mathbf{C},J)$ and $(\mathbf{D},K)$ such that $Sh(\mathbf{C},J)\simeq Sh(\mathbf{D},K)$. 
Is this still the case when $\mathbf{C} \simeq \mathbf{D}$? That is, is it possible to have distinct sites $(\mathbf{C},J)$ and $(\mathbf{C},K)$--note the same underlying category--such that $Sh(\mathbf{C},J)\simeq Sh(\mathbf{C},K)$?
 A: Yes, they can, and this observation is fundamental to Olivia Caramello's bridge technique.
The comparison lemma gives a sufficient condition that the resulting toposes agree. For instance, let $X$ be a topological space and let $\mathcal{B}$ be a basis for $X$ (a set of open subsets such that any open subset is a union of open subsets of $\mathcal{B})$. Then there is the canonical site associated to $X$, having all open subsets of $X$ as objects. (This is just the site implicitly used when defining the sheaf topos $\mathrm{Sh}(X)$.) You could also define a subsite which has only the open subsets contained in $\mathcal{B}$ as objects. The comparison lemma guarantees that the resulting sheaf toposes will be canonically equivalent.
An example where the sites are not so obviously related and the resulting sheaf toposes are still equivalent is the following. Let $R$ and $S$ be Morita-equivalent rings (that is, their categories of (left, say) modules are equivalent). (Sometimes ring theorists say that the matrix ring $R^{n \times n}$ is not "seriously noncommutative", since it is Morita-equivalent to $R$.) Then the classifying toposes of the theory of $R$-modules and of $S$-modules will be equivalent. Their underlying sites, the syntactic sites of the two theories, will look quite different (unless $R$ and $S$ happen to even be isomorphic).
In the spirit of this example, sites which yield equivalent categories of toposes are also called "Morita-equivalent".
