Does this show that equality (set theoretically) does not necessarily imply isomorphic (categorically)? I was thinking about a rather simple question, namely: If two things are equal, are they necessarily isomorphic? At first, I thought this was a silly question as it seems that the identity map should provide the isomorphism we are after. However, after thinking about it a little bit more, I started to think of what does equality and isomorphism mean. Thus, I decided to work in a case where I know what equality and isomorphism mean. Here when I say equality, I mean equality of sets (i.e. both sets are subsets of each other), and when I say isomorphism, I mean that in the category that we are working given two objects $A$ and $B$ they are isomorphic if there is $f:A\rightarrow B$ and $g:B\rightarrow A$ such that $f\circ g=1_B$ and $g\circ f=1_A$. Thus, I thought that if we can cook up a category where $A=B$, but there are no morphisms between $A$ and $B$, then we must have that $A=B$, but $A\not\cong B$.
Thus, I consider the following category where the objects in the category are sets and the morphisms are functions of sets. In this category, I will take two distinct objects $A$ and $B$ where $A$ is the empty-set and $B$ is also the empty set, and the only morphisms will be the identity morphisms $1_A$ and $1_B$. Thus, if we think of the associated graph, it will just consist of two nodes with self loops. Now clearly, we must have that $A=B$ and $\emptyset\subset\emptyset$, but $A\not\cong B$ since there are no arrows between $A$ and $B$. Thus, would this constitute an example where equality does not imply isomorphism?
Furthermore, one might say that this is a silly example since there is a larger category that has this as a subcategory where $A\cong B$ (just "add the identity map between the objects"). Thus, is it always the case that if I have a category where there are objects which are equal but not isomorphic to just "add the identity morphism" between them to arrive at a larger category where equal objects are now isomorphic and contains the original category as a subcategory?
 A: When you define a category, you need a set (or class) of objects $\text{Ob}$. If your objects are themselves sets, $A,B\in\text{Ob}$ and $A = B$ as sets, then they are indistinguishable as objects.
A: As has already been said, the most common way to define a category is as having a class of objects, and classes cannot have duplicates of the same element. It's also possible to imagine an approach in which the collection of objects in a category does not come equipped with an equality predicate at all--that is, it's meaningless to ask whether two objects of a category are equal. This is, in fact, a more categorically natural approach, though it doesn't sit very well with the standard "everything is a set" foundation. In this case, the question "does isomorphism imply equality" is meaningless, rather than having a positive answer. In principle you could imagine defining a "category" with a multiset of objects, exhibiting the behavior in your post, but then you would simply have proven that you aren't really talking about categories as usually understood at all.
A: No, you can't do this. The reason is that the fundamental characteristic of equality---whatever equality is taken to actually mean---is that equal things may be substituted for one another in a way that the rest of the theory must respect.
So, for instance, it is completely fine to consider a theory where there are multiple, distinct empty sets. This is not how set theory is usually defined, because the axiom of extensionality declares all empty sets to be equal, but you could remove extensionality. However, once you do say that two sets $A$ and $B$ are equal, you can't go on to say that $A \cong A$, but not $A \cong B$, because if $A = B$, I must be allowed to substitute $B$ for $A$ on the right of $A \cong A$. And the only way we could have $A \ncong B$ is if $A \ncong A$ also, for the same reason.
So, there is plenty of room to negotiate about which things are considered equal. Maybe sets are equal when they were defined by the same formula. Maybe they're equal when they have the same elements. Maybe they're equal when they have the same number of elements. But, the thing that is not negotiable is that however equality is characterized, the rest of the theory must somehow be invariant with respect to interchanging equal things. Otherwise you are not really talking about what people mean by, "equality."
