Hello, I remember that the calculation of the self isomorphism is found by symmetry and rotation and I followed this calculation to calculate the self isomorphism of this diagram but answer is wrong. Am I missing some diagram? Or did I make mistakes in the calculation? Here are my thoughts on the calculation. I think this triangle has 6 automorphisms, plus rotations, so this diagram has 6*4=24.
The vertices of a triangle in this graph cannot be permuted arbitrarily. This already follows from the fact that two of the vertices have degree $2$ and the other has degree $3$.
The graph has exactly one vertex of degree $4$ in its middle, which means that this vertex is fixed under any automorphism.
Then the four neighbours of the middle vertex can at most be permuted. And as the graph is clearly symmetric in this respect, they actually can be permuted arbitrarily, giving rise to a factor $4!$.
Once you have decided how to permute these four vertices, the two other neighbours of each of these can at most be permuted (and they actually can). This gives rise to a factor $2!$ -- for each of the four triangles.