Affine Plane of Order 4 Picture?

Solution 1:

Start with a finite field of order $4$:

$$\begin{array}{c|cc} + & 0 & 1 & a & a+1 \\\hline 0 & 0 & 1 & a & a+1 \\ 1 & 1 & 0 & a+1 & a \\ a & a & a+1 & 0 & 1 \\ a+1 & a+1 & a & 1 & 0 \end{array}\qquad \begin{array}{c|cc} \times & 0 & 1 & a & a+1 \\\hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & a & a+1 \\ a & 0 & a & a+1 & 1 \\ a+1 & 0 & a+1 & 1 & a \end{array}$$

Now you can draw the points as a $4\times4$ grid using these elements as $x$ and $y$ coordinate. That's the $n^2$ points you'd expect from an affine plane of order $n$. You can also draw lines of constant $x$ coordinate (i.e. vertical lines), and non-vertical lines with equations $y=tx+u$ where $t$ and $u$ are again elements from your field. This is $n+n^2$ lines, as expected.

It doesn't really matter how you draw the lines, as long as the points on each line come from the same equation. If the lines you draw intersect in points besides the 16 points in the plane, that simply doesn't count as an intersection. So one possible illustration would be this:

Here I've drawn lines of the same “slope” (i.e. same parameter $t$) in the same color. Cyan is $t=0$, navy blue is $t=1$, magenta is $t=a$ and dark green is $t=a+1$. The bright green lines are vertical, $t=\infty$ if you will.

Solution 2:

In this picture, opposite red points are the same. For example, the vertical line appears to contain 5 points, but the 2 red points are the same, so it just contains 4 points.

Solution 3:

The dotted rectangle represents a wrap-around region, so green lines leaving one side come back in on the opposite side:

Intersections and Parallels
Each of the 15 green lines passes through 4 points, as does each of the 5 blue lines. Each green line is parallel to (i.e., has no point in common with) the two that are vertically offset from it, as well as to one of the blue lines. It is easy to verify that the center point and any other point determine a unique line, and so by symmetry the same is true for all 15 points in the dotted box. Finally, any two lines clearly intersect in at most one point, and every point is on five lines, so this is indeed the affine plane of order 4.

Symmetries
This picture makes it look like the bottom vertex is special, but of course any of the 16 points can be placed at the bottom. The 3-way symmetry and 5-way symmetry of the remaining 15 points is especially clear here, since the three rows or five columns can be rotated without changing the picture. We can also see a 4-way symmetry (3 length-4 cycles and a length-2 cycle) involving 14 points: We can permute the columns, moving column $i$ to column ($2i$ mod $5$), and swap two rows, and the picture will look the same. Reversing the columns corresponds to doing this permutation twice.

There is one final symmetry that is not very clear from this diagram, that involves 12 points. This paragraph will describe this symmetry. If we keep one line fixed (say the blue line on the left), then the remaining 12 points can be arranged in 12 different ways, allowing any point to be in any position (but then the other 11 points' positions are forced). Note that these 12 points lie on three green parallels to the fixed blue line. We can permute their 4 blue columns with any even permutation (the four points on each parallel staying within that parallel), but we must then also vertically rotate the 12 points (moving each parallel onto the next) according to the even permutation used. If we imagine a tetrahedron whose 4 corners are associated with the 4 columns, then an even permutation of the corners corresponds to either (A) a 0° rotation or 180° rotation (2 opposite sides are each reversed), (B) a 120° rotation (a face is rotated) clockwise, or (C) a 120° rotation counter-clockwise. There is no vertical rotation of the 12 points for case (A), while cases (B) and (C) correspond to vertically rotating up or down. If we think of the tetrahedron as having rotatable faces (like a Rubik's cube), then we can permute the 4 columns by rotating the faces, but each rotation also rotates the 12 points vertically. To put a point where you want it, put it in the correct column and then rotate the opposite face (the one that doesn't contain the corner for that column) to put the point at the desired height. The twelve possible orientations of the tetrahedron correspond to the twelve possible arrangements of the twelve points, and each point is clearly at a different position for each of the twelve orientations. (As you might suspect, it is also possible to associate the 12 points with the vertices of a truncated tetrahedron so that the rotations of the solid object correspond to the ways the points can be rearranged. Thinking of it that way is cleaner in some ways, but I find the relation to the above diagram less obvious that way.)

Comparison with Pentagram Diagram
The $5\times3$ symmetry is the same as is visible in the pentagram diagram, in which the largest pentagon is the same as the smallest, so the 3-way symmetry corresponds to cycling through the sizes, with a 180° rotation between levels. The blue lines here correspond to the lines through the center in the pentagram diagram.