I don't know.  The following is a near miss which might be useful. 

Start with a hexagonal cycle path ABCDEFA.  Duplicate point C to C' and connect C' to
B,C, and D.  Similarly duplicate points E and F, and add edges EE', FF', and the 3 edges
to form the path DE'F'A.  Then it has diameter 3, but the only point that is distance 3 from E (and also from E') is B, so it cannot accomdate such a permutation.  The only problem is that vertex D has degree 4, so the graph is just shy of being 3-regular.

It may be possible to use this by stitching together two large even cycles to get
a regular graph (with the property that two vertices must share an antipode), but I will let someone else do it.

**EDIT 2011.05.12**
Thanks to Alain, Aaron, and Roland for their encouragement and checking, I will bring
the comment's example into this answer.  Indeed two 12-cycles can be stitched together,
say at vertices 3 and 4, and at 6 and 7, and at 10 and 11, with 6 vertices and edges added to form 3 separate ladders as in the example above.  This time the graph is 3-regular, and
points 1 and 1' have 7 as the unique common antipode, as do also 9 and 9' share vertex 3 as an antipode.  The result has 27 edges, 18 vertices, and diameter 6, and does not admit a permutation that takes every vertex to one at distance 6 from that vertex, because e.g. there are not enough antipodes for 9 and 9' to share in such a permutation.  

The "ladders" in the construction can be replaced by graphs which have something like a complete graph on k vertices at each rung (instead of the complete graph on two vertices as in the present example) to get examples with arbitrarily high regularity and in which k vertices share an antipode.
**END EDIT 2011.05.12**

Gerhard "Cycles Can Make Me Dizzy" Paseman, 2011.05.11