More Symmetrical

We will approach this problem exhaustively.

For a square, there are only 4 possible symmetries, which are vertical, horizontal, top left to bottom right, and top right to bottom left. On the $5 \times 5$ grid, the possible lines of symmetry look like this:

Now, if a grid has both verticle/horizontal and diagonal symmetry, then the grid has all 4 symmetries. To prove this, let's put the grid on a coordinate axis with its center at $(0,0)$ , like so:

Now, let us construct a grid that has vertical and top left to bottom right symmetry. On our coordinate plane, this means that the grid is symmetrical over the line $x=0$ (vertical symmetry) and the line $y=-x$ (diagonal symmetry). If we have a point on our grid $(x,y)$ which is shaded in, then by vertical symmetry (a reflection over the line $x=0$ ) the point $(-x,y)$ must also be shaded in on our grid. Applying diagonal symmetry (a reflection over the line $y=-x$ ) results in two new points: $(x,y)$ gives $(-x,-y)$ and $(-x,y)$ gives $(x,-y)$ . We now have 4 points. The result looks like this:

In this picture, the original point is U, and the 3 other points are the result of reflections. This grid obeys all four symmetries, therefore if a grid obeys a vertical/horizontal symmetry and a diagonal symmetry, it obeys all four symmetries. This means that there are only 7 possible combinations of lines of symmetry: vertical, horizontal, top left to bottom right, top right to bottom left, top left to bottom right AND top right to bottom left, vertical AND horizontal, and all 4 lines of symmetry.

Since we want the grid to obey at least 2 symmetries, the symmetries must be one of the following: top left to bottom right AND top right to bottom left, vertical AND horizontal, and all symmetries. We test the three possible cases:

Case 1: Symmetry along both diagonals:

3 squares added.

Case 2: For vertical AND horizontal symmetry:

5 squares added. Note that because the grid already had diagonal symmetry, adding vertical symmetry results in a grid which has all four symmetries, meaning we don't have to test the case of all 4 symmetries.

The answer is therefore 3, by exaustion.