This is the net of a square-based pyramid. How many such nets are there in total, not counting those which are equal by rotations/mirror symmetry?

The question you're asking is a classic problem in geometric combinatorics. A net of a polyhedron is a pattern that you can cut out and fold to form the polyhedron. It's a way of representing three-dimensional shapes in two dimensions.

For a square-based pyramid (a pyramid with a square base and four triangular faces), there are 11 distinct nets. This was proven mathematically in a paper by G. N. Frederickson and J. O'Rourke in 1997.

Here's a rough idea of how you might come to that conclusion:

1. Start by considering the square base. There are four triangular faces that must be attached to it.

2. There are four edges to the square, so you might think that you could attach one triangle to each edge. But that doesn't work, because then you wouldn't be able to fold it up into a pyramid. The triangles would overlap.

3. So at least one edge of the square must be connected to two triangles. That gives you a few different possibilities to explore.

4. For each possibility, you then have to consider how the remaining triangles can be attached to the other faces. This is where it gets tricky, and you have to be careful not to count the same net twice.

5. After going through all the possibilities, you find that there are 11 distinct nets.

Remember, this is a simplified explanation. The actual proof involves more detailed geometric reasoning.