The general approach is first to find specific points that must be fixed under an automorphism, then proceed inductively using the following sorts of logic:
- If a node is known to be fixed, and all but one of its neighbors are known to be, then we know that all neighbors must be fixed.
- If two nodes are known to be fixed, and there is only one simple path of length $k$ between those two nodes, then all the nodes on that path must be fixed.
- If two neighboring nodes are fixed, and there is only one cycle of length $k$ containing those two nodes and the edge between them, then all the nodes on that cycle must be fixed.
First part: Show specific nodes are fixed
The set of nodes not on any triangle is $\{3,7,10\}$. Any automorphism must restrict to an automorphism of the sub-graph on these three nodes. In particular, it must send $3$ to $3$, and thus $3$ must be fixed.
There is only one edge between $3$ and a node on a triangle. Since $3$ is fixed, that edge must be fixed, and therefore it's other end must be fixed. So $11$ must be fixed.
Second part: Induction
Since $3$ and $11$ are fixed, and there is only one cycle of length $4$ containing the edge $\{3,11\}$, then the other nodes of that cycle, $6$ and $10$, must be fixed. (Rule 3.)
Since $3$ and two neighbors $11,10$ are fixed, $7$ must be fixed (Rule 1.)
Since $6$ and $11$ are fixed, and there is only one cycle of length $3$ containing the edge between them, the third point on that cycle, $2$, must be fixed. (Rule 3.)
Since $7$ and $10$ are fixed, and there is only one path of length $3$ between them, the other points on that path, $0,4$ must be fixed. (Rule 2.)
Since $0,4$ are fixed, $8$ must be fixed. (Rule 3.)
Since $7,0,3$ are fixed, $1$ is fixed. (Rule 1.)
Since $1,8$ fixed, and there is only one path of length $2$ between them, the point in between, $5$ must be fixed. (Rule 2.)
Finally, $9$ is fixed because all the other nodes are fixed.