Show that given matrices are not similar $$A=\begin{bmatrix} 0 &1 &0 &0 \\0 &0 &0 &0 \\0 &0 &0 &0 \\0 &0 &0 &0 \end{bmatrix}$$
$$B=\begin{bmatrix} 0 &1 &0 &0 \\0 &0 &0 &1 \\0 &0 &0 &0 \\0 &0 &0 &0 \end{bmatrix}$$
How they are or are not similar. We have that the characteristic polynomial for both $A$ and $B$ is $x^4$ with the minimal polynomial of $x^2$.
On
Suppose that $A = P^{-1}BP$ for some invertible $P$. Then, $A^2 = P^{-1}B^2P$, but $A^2 = 0$, so $P^{-1}B^2P = 0$ which implies $B^2 = 0$, but this is not true, as you can check.
One can also use the fact that they have dissimilar rank to see this, since if vectors $c_i$ span the columns of $A$ then $P c_i$ span the columns of $B$ for some invertible $P$.
The key point in your example, as you have already pointed out, is that the minimal polynomial and the characteristic polynomial are the same for these two matrices, but they are not similar. While two matrices may share many properties such as trace, determinant, characteristic polynomial etc. one must compute their rational canonical form to see if they are similar to each other or not.
Hint: similar matrices have the same rank.