Let the following matrix be given. Note that we are in the field consisting of five integers:
F = (0, 1, 2, 3, 4)
$A = \begin{bmatrix} 1 & 2 & 0\\ 3 & 2 & 1\\ 0 & 2 & 2 \end{bmatrix}$
The Jordan Normal Form of the matrix is
$J = \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix}$
Now, to find the basis transformation matrix, I know first of all that $A^3 = 0$. Now I want to consider $A^2 = \begin{bmatrix} 2 & 1 & 2\\ 4 & 2 & 4\\ 1 & 3 & 1 \end{bmatrix}$
Choosing $v_3 = \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$, we see that $A^2v_3 = \begin{bmatrix} 2\\ 4\\ 1 \end{bmatrix}$, so $v_3$ is not in the kernel of $A^2$. I will set $v_2 = \begin{bmatrix} 1 & 2 & 0\\ 3 & 2 & 1\\ 0 & 2 & 2 \end{bmatrix} * \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} = \begin{bmatrix} 0\\ 1\\ 2 \end{bmatrix}$ and $v_1 = \begin{bmatrix} 1 & 2 & 0\\ 3 & 2 & 1\\ 0 & 2 & 2 \end{bmatrix} * \begin{bmatrix} 0\\ 1\\ 2 \end{bmatrix} = \begin{bmatrix} 2\\ 4\\ 2 \end{bmatrix}$
Now set $S = [v_1, v_2, v_3] = \begin{bmatrix} 2 & 0 & 0\\ 4 & 1 & 0\\ 2 & 2 & 1\\ \end{bmatrix}$. I now want to find the inverse of S so I can check whether $S^{-1}*A*S = J$, but I am not sure how I can do this while staying in the field F, as fractions inevitably come up when I do it normally. I am also not entirely sure if my approach to finding S was correct.
Thank you for your help.
Your approach to finding $S$ is correct.
The fact that $A^2 \neq 0$, $A^3 = 0$, and $A^2 v_3 \neq 0$ ensures that the vectors $A^2v_3, Av_3, v_3$ are linearly independent over the field $\Bbb F$. As such, the matrix $S$ that you construct has linearly independent columns, and so it is necessarily possible to compute its inverse while staying in the field $\Bbb F$. The trick is that instead of dividing as you normally would (which would lead to fractions), you multiply by multiplicative inverses.
Let's compute the inverse of your matrix $S$ using Gaussian elimination. We reduce the augmented matrix $[S \mid I]$, $$ \pmatrix{2 & 0 & 0 & 1 & 0 & 0\\ 4 & 1 & 0 & 0 & 1 & 0\\ 2 & 2 & 1 & 0 & 0 & 1}. $$ To begin, we could "divide" the first row by $2$. Note that "dividing" in this context means multiplying by the multiplicative inverse. In other words, the number that we should multiply the first row by is the solution to the equation $2 x \equiv 1 \pmod 5$. You can verify that $x = 3$ is the solution, since $2(3) = 6 \equiv 1 \pmod 5$. So, our first step will be to multiply the first row by $3$. This leads to $$ \pmatrix{1 & 0 & 0 & 3 & 0 & 0\\ 4 & 1 & 0 & 0 & 1 & 0\\ 2 & 2 & 1 & 0 & 0 & 1}. $$ As it so happens, no further divisions are required now. For the next operations we can subtract $4R_1$ from $R_2$, $2R_1$ from $R_3$, and $2R_2$ from $R_3$. Doing so and reducing entries modulo 5 along the way results in the reduced matrix $$ \pmatrix{ 1 & 0 & 0 & 3 & 0 & 0\\ 0 & 1 & 0 & 3 & 1 & 0\\ 0 & 0 & 1 & 3 & 3 & 1}. $$ So, the inverse of $S$ in $\Bbb F$ is $$ \pmatrix{3 & 0 & 0\\3 & 1 & 0\\3 & 3 & 1}. $$