Erik Demaine asks:

<<
Can't any two topologically equivalent embeddings (in the sense that the
cyclic order around each vertex is identical in two embeddings) be obtained
by continuous deformation? 
>>

This looks fairly easy to prove.  And I'm fairly convinced of John's assertion that a  straight-line embedding of a graph G in the plane *can* be obtained by a continuous deformation of an arbitrary planar embedding of G.

But given that a graph G that can embedded in the plane can be embedded with all edges straight, however, doesn't immediately imply that the straight-line embedding *is* topologically equivalent to the original embedding (even though the graph itself is of course topologically the same).

For example consider a square with a 5th node making a triangle with one side of the square -- a graph with 5 nodes and 6 edges.  This can be embedded in the plane with the 5th node either inside or outside the square.  (Of course, straightenability is not an issue here.)

--Dan