In my project to construct the outer automorphism group for $S_6$ I have come across a need (or desire) to visualize a graph that has 15 vertices, with each vertex having 6 incident edges and 3 incident "triangles." (By my count, that makes 15 total vertices, 45 total edges, and 15 total triangles.) No matter how I try to draw this, I am not successful. I started with a regular 15-sided figure, but this lead nowhere quickly, and I can see clearly that if this graph even exists, it will have to be a concave/self-intersecting polygon, geometrically. Am I barking up the wrong tree, or is there a way to realize this graph?
Edit: In case anybody is wondering, the reason I am interested in this graph is because I am considering all the two-cycles in $S_6$ as "vertices," for which there are $6 \choose 2$ aka 15 total vertices; two vertices are joined by an edge if they commute, ie, if they are disjoint. Three vertices form a "triangle" if they are pairwise disjoint.
Start with $15$ disjoint triangles, and identify the vertices in sets of $3$ (belonging to distinct triangles, but making sure that you don't identify all the vertices of one triangle with those of another) so that you end up with $15$ distinct vertices.
FWIW, here are some pictures of the graph as constructed using Derek Holt's comment. The graph is not planar, and I don't know if there is a simpler way to visualize it.