Uniquely 4-colorable Unit Distance Graphs A graph is "unit distance" if it can be embedded in the plane 
in such a way that whenever two vertices are connected by an edge, 
their distance is $1$. 
A graph is uniquely $4$-colorable if there is a unique way 
(up to renaming of colors) of coloring its vertices with $4$ 
colors in such a way that no two adjacent vertices have the 
same color, and if no such coloring with $3$ colors exists.
Question: Is there finite graph that is both uniquely 
$4$-colorable and unit distance? 
 A: EDIT the original answer was wrong, thanks to Jan Kyncl for pointing the mistake, hopefully it is fixed now.
I think this is open, since existence will improve the lower bound of chromatic
number for unit distance graphs from $4$ to $5$ almost surely.
Assume uniquely $4$ colorable finite unit distance graph $G$ exists. Color it
with colors $a,b,c,d$. Take two copies of $G$: $G_1$ and $G_2$. For vertices
colored $a$, $a_1 \in G_1, a_2 \in G_2$ merge $a_1$ and $a_2$ to vertex $A$ to
get unit distance graph $G'$ with the property that all vertices in $G_1,G_2$ which are colored $a$ have the same color in all $4$ colorings of $G'$. For vertices $a_1' \in G_1,
a_2' \in G_2$ colored with the $a$ color, rotate $G_2$ with center $A$ trying
to get $a_1',a_2'$ at distance $1$. The triangle inequality is enough. If you
can do this, add edge $(a_1',a_2')$ to get unit distance graph, which is not $4$
colorable. I think such four vertices exist with high probability and comment by Pat Devlin confirms this.
