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

1$\begingroup$ It's a monoid! I mean, it's equivalent to the endomorphism monoid of any object. $\endgroup$ – Fernando Muro Dec 25 '18 at 18:22
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 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 oneobject categories.
A category with one object is the same thing as a 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 2category, then the full sub2category of categories with one isomorphism class of objects is biequivalent to the full sub2category of categories with one object, which is biequivalent to the 2category 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 $g$.

$\begingroup$ Thanks Tim. If our category is abelian, can one say something interesting about it( of course in this case all non zero objects are isomorphic, the category will have two objects) $\endgroup$ – Diego Havez Dec 25 '18 at 4:13

1$\begingroup$ An abelian category has a zero object, so if all objects are isomorphic, then the category is trivial (it is equivalent to the terminal category, i.e. the oneobject category with no nonidentity morphisms). However, a preadditive category (i.e. a category enriched in abelian groups) with one object is the same thing as a ring, and a preadditive category with one isomorphism class of objects is additively equivalent to a preadditive category with one object. $\endgroup$ – Tim Campion Dec 25 '18 at 4:18

$\begingroup$ Yes but I said all nonzero objects are isomorphic.In this case we have two classes. Zero class and another one. $\endgroup$ – Diego Havez Dec 25 '18 at 4:20

$\begingroup$ As you observe, you can again take a skeleton, so if all nonzero objects are isomorphic (and there is a nonzero object) then your category is equivalent to a category with a unique nonzero object $X$. But abelian categories also have direct sums. So we will have $X \cong X \oplus X$. An abelian category also has kernels and cokernels, so there will be lots of new objects you can construct which must be isomorphic to $X$ as well. I don't know an example of such a category. $\endgroup$ – Tim Campion Dec 25 '18 at 4:25

3$\begingroup$ You're welcome. I suppose you could start with something like the category of countabledimensional vector spaces over field $k$, and quotient by the Serre subcategory of finitedimensional vector spaces. That ought to give an example. But it seems like a strange condition to me. $\endgroup$ – Tim Campion Dec 25 '18 at 4:29