This was essentially answered in your first question, it was only obscured because the example I gave happened to be uniquely colorable. But here's a way to construct lots of other examples.
Start with a triangulation in which all degrees are even. (For example, let's start with the octahedral graph.)
Draw a line between two non-adjacent vertices and add a vertex wherever this line crosses an existing edge. This (always!) gives us a triangulation in which exactly two vertices are odd:
Those vertices, as we know, must have the same color in any $4$-coloring. However, usually such a triangulation will not be uniquely $4$-colorable, and in fact that's not true here.
Finally, just add a new vertex adjacent to the three corners of some face (any face):
There are now more than two odd vertices (no matter which face we picked). However, any $4$-coloring of this graph induces a $4$-coloring of the graph at the previous step, and vice versa (the color of the new vertex is uniquely determined). Therefore:
- The graph we get isn't uniquely $4$-colorable, because we didn't have a uniquely $4$-colorable graph before.
- The two vertices that used to be the only two odd vertices before must still have the same color in any coloring.