"Categorical" Schröder–Bernstein theorem?

The Bernstein-Schröder theorem also fails for posets : for example, take $A$ to be the poset of natural numbers and $B=A\sqcup A$, with no order relation between the two copies of $A$. $A$ can be embedded into $A$ as either copy, $B$ can be embedded into $A$ by identifying one copy with the odd numbers and the other with the even numbers. The two posets are not isomorphic since $A$ is a total order and not $B$. Note that this implies that as categories, they are not even equivalent.

This MO answer gives another counterexample : the posets $[0,1]$ and $[0,1)$ can be embedded into each other, but are not isomorphic. Here the two orders are total, and the embeddings are initial segments. This means that as functors between categories, they are actually fully faithful; yet the two corresponding categories are, again, not even equivalent since only one has a terminal object.

The analogue of Schröder–Bernstein is false for groups, as shown here. Let $f:G\to H$, $k:H\to G$ be such an example. We can translate this example to the case of categories as follows.

Let $\mathbf BG$ be the category with one object, called $\bullet$, and with $\mathrm{Hom}(\bullet,\bullet)=G$. Composition of morphisms is by multiplication in G, and $\mathrm{id}_\bullet$ is just the identity of $G$. Define $\mathbf BH$ analogously.

Now define a functor $\mathbf Bf:\mathbf BG\to \mathbf BH$ by sending $\bullet_G$ to $\bullet_H$ and acting on morphisms by $f$. Likewise define $\mathbf Bk$.

Then $\mathbf Bf$ and $\mathbf Bk$ are injective on objects and morphisms, but $\mathbf BG$ and $\mathbf BH$ aren't isomorphic because $G$ and $H$ aren't.