Conjugate vertices in a graph1 or conjugate elements of a group2 are equivalent (indistinguishable, essentially the same) in one specific structural sense.
Isomorphic objects in a category are equivalent in another specific structural sense.
In both cases we don't look inside the objects but declare them equivalent from the outside.
In both cases equivalence has to do with isomorphism: with structure preserving maps between the structure the conjugate elements live in and itself (automorphisms) resp. with iso arrows between the elements themselves.
Define two objects $A,B$ in a category to be conjugate when there is an isomorphism endofunctor $F$ with $F(A) = B$.
Question 1: Is it true that any two isomorphic objects are conjugate (since there is an isomorphism endofunctor that permutes them)?
The reverse is most certainly false: There are categories with conjugate objects that are not isomorphic. E.g. the graphs
alt text http://epublius.de/mathoverflow/graph6.png
in the category of graphs over two fixed vertices (with graph homomorphisms as morphisms) are two such objects (#9 and #6 in the diagram below). Note that there is no morphism at all between these two graphs.
Question 2: Might it be the case that whenever two objects are conjugate-but-not-isomorphic there is no morphism between them? Or is this true only in special categories and/or special cases?
Question 3: How "normal" is it that a category contains conjugate-but-not-isomorphic objects?
Most of all I'd like to know how to think about this bewildering pair of equivalences in general terms.
Appendix
Here is the complete category of graphs over two fixed vertices and an arrow whenever there is a graph homomorphism. Compositions and identities are omitted.
alt text http://epublius.de/mathoverflow/2graphcategory_small.png
The numbers are derived from the adjacency matrices: 0 = 00|00, 1 = 10|00, ..., 15 = 11|11.
Footnotes
1 $x,y$ are conjugate iff there is a $g \in \text{Aut}(G)$ with $g(x) = y $.
2 $x,y$ are conjugate iff there is a $g \in G$ with $gx= yg$.