Conjugate vertices in a graph<sup>1</sup> or conjugate elements of a group<sup>2</sup> 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][1x] [<sup>(source)</sup>][1]
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
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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][2x] [<sup>(source)</sup>][2]
The numbers are derived from the adjacency matrices: 0 = 00|00, 1 = 10|00, ..., 15 = 11|11.
The numbers of the two graphs above are 10|01 = 9 and 01|10 = 6.
Footnotes
----------
<sup>1</sup> $x,y$ are conjugate iff there is a $g \in \text{Aut}(G)$ with $g(x) = y $.
<sup>2</sup> $x,y$ are conjugate iff there is a $g \in G$ with $gx= yg$.
[1]: http://epublius.de/mathoverflow/graph6.png
[2]: http://epublius.de/mathoverflow/2graphcategory_small.png
[1x]: https://i.sstatic.net/oUp6n.png
[2x]: https://i.sstatic.net/nuAVN.png