Isomorphism is the correct notion of equality between objects in a category. From the category-theoretic point of view, if you want to distinguish between two objects which are isomorphic but not equal, it means that the morphisms in your category don't preserve whatever aspect of the objects allows you to make this distinction, and hence the category doesn't really capture what you want to be working with. If you want to talk about it categorically, you should consider a category with morphisms that preserve all of the structure you care about, including whatever allowed the distinction to be made.

For example (this example is due to Qiaochu Yuan), consider the category with objects metric spaces, and morphisms continuous maps. The diameter of a metric space , which is the maximum value of $d (x, y)$ for $x, y \in X$ , is a feature of metric spaces which is not invariant under isomorphism in this category; for example, the subsets $[0, 1]$ and $[0, 2]$ of $R$ , equipped with the usual metrics inherited from $R$ , are isomorphic in this category. There is a continuous map $f : [0, 1] \to [0, 2]$ and a continuous map $g : [0, 2] \to [0, 1]$ such that $f g$ and $g f$ are identities. For example, one could take $f$ to be multiplication by $2$ , and $g$ to be division by $2$ . However, the diameter of $[0, 1]$ is $1$ , and the diameter of $[0, 2]$ is $2$ . Therefore, insofar as "diameter" is a property of metric spaces, the objects of these categories are not metric spaces. The correct name for them is "metrizable spaces", since this category is equivalent to the category whose objects are topological spaces whose topology is induced by some metric and whose morphisms are continuous maps.

For a less realistic (but more obvious) example, consider the category of groups and arbitrary functions between their underlying sets. The objects of this category are, supposedly, groups, but properties of groups, such as "simple", do not respect isomorphism in this category.

Another example of this is the product of, say, sets. It determines a functor $Set \times Set \to Set$ . We would like to say that this is associative, but this is false; a typical element of $A \times (B \times C)$ looks like $(a, (b, c))$ , while a typical element of $(A \times B) \times C$ looks like $((a, b), c)$ . Since these sets have different elements, they are not equal. However, they are isomorphic. In fact, the two functors $Set \times Set \times Set \to Set$ given by $(A, B, C) \mapsto A \times (B \times C)$ and $(A, B, C) \mapsto (A \times B) \times C$ are isomorphic in the category of functors $Set \times Set \times Set \to Set$ . That is, they are naturally isomorphic.

link to a "natural isomorphism" page