I'm assuming that the reverse of a cyclic ordering counts as the same ordering. I'm also assuming the graph is simple (otherwise subdivide edges with extra vertices to make it simple).

There is only one cyclic order for vertices of degree 1,2,3, so the interesting vertices have degree 4 or more.

3-connected graphs have only one planar embedding. Given any planar embedding of a connected graph, you can get all the other planar embeddings by reordering and flipping over the parts separated by a 1-cut or a 2-cut.

So, without thinking about it too much, I think the answer is:

A vertex has consistent ordering if it has degree 1,2,3, or if it is not a cut-vertex or part of a 2-vertex cut.

A cut-vertex of degree at least 4 does not have consistent ordering.

The remaining case is a vertex $v$ of degree at least 4 that is part of a 2-vertex cut. Separate the graph at the cut so that $v$ and the other vertex in the cut are replicated in each piece (I hope this is clear enough without a formal definition). FIXED: Then $v$ has consistent ordering iff there are exactly two pieces and in one of them $v$ has degree 1.