TITLE: Showing that a specific regular graph has a trivial automorphism group
QUESTION [6 upvotes]: Consider the graph $G$ in the following picture 
 
It can be verified (using Sage or an equivalent program) that $G$ has a trivial automorphism group.
What I am wondering is how to show this fact by a formal (by hand) proof that uses the least amount possible of case analysis. Anyone happens to see a short proof?

REPLY [2 votes]: Just say something special about each vertex. Remove the vertices participating in a triangle. You are left with 7-3-10 whence 3 stays fixed. 11 is the only neighbor of 3 participating in a triangle, so 11 stays fixed. 11-10-6-3 is the unique shortest chain from 11 to 3 that is not just the edge joining them, so 10 and 6 are fixed and thereby 7 stays fixed. So now it is 3,6,7,10,11. 11-2-9-1-7 is the unique shortest chain from 11 to 7 not involving the vertices we fixed already, so 2,9, and 1 cannot move. Similarly, 10-4-0-7 is the unique shortest chain from 10 to 7 bypassing the already fixed vertices, so 0,4 stay put.
Now we have only 5 and 8 left and about 100 reasons not to be able to swap them.
Still boring, but also no casework anywhere.