What one can say about a category in which, any two objects are isomorphic. I know that this is a strange question.
Thanks



Category with one isomorphism class
The question in the title and the body are slightly different -- a category with all objects isomorphic either has one isomorphism class of objects, or *zero* isomorphism classes of objects (i.e. it could be empty). Probably the right class of categories to consider is the one from the title -- categories with one isomorphism class of objects, i.e. nonempty categories with all objects isomorphic.


Let $C$ be a category. If $C$ is nonempty and all the objects of $C$ are isomorphic, then the [skeleton](https://ncatlab.org/nlab/show/skeleton) of $C$ is a category with one object. Conversely, of course, if $C$ has one object, then all of its objects are isomorphic. So up to equivalence, nonempty categories with all objects isomorphic are the same as one-object categories.


A category with one object is the same thing as a [monoid](https://en.wikipedia.org/wiki/Monoid). In one direction, if $C$ is a category with one object $\bullet$, the homset $Hom\_C(\bullet, \bullet)$ is a monoid. Conversely, if $M$ is a monoid, then there is a category $BM$ with one object $\bullet$ a and $Hom\_{BM}(\bullet,\bullet) = M$, with composition given by multiplication in $M$. This correspondence is an equivalence of categories between the category of categories with one object and the category of monoids.


There are other things one could say, but that's the long and short of it. Maybe I'll mention that if you consider $Cat$ as a 2-category, then the full sub-2-category of categories with one isomorphism class of objects is biequivalent to the full sub-2-category of categories with one object, which is biequivalent to the 2-category of monoids, homomorphisms, and intertwiners. Here, if $f,g: A \to B$ are monoid homomorphisms, an intertwiner $\beta: f \Rightarrow g$ consists of $\beta \in B$ such that for all $a \in A$, $\beta f(a) = g(a)\beta$. For example if $A,B$ are groups, then an intertwiner exhibits $f$ as *conjugate* to $g$.


