I am concerned on this interesting question
Given matrix $A$, does the Jordan cardinal form have the most zeros, in other word, it has the least nonvanishing indices, among the equivalent class of similarity.
I can prove this argument for diagonalizable matrices and nilpotent matrices as follow. Now fix $n$, the size of $A$.
If $A$ is diagonalizable, then its Jordan cardinal form is nothing but diagonal matrix. Thus the gross number of zeros is $n^2-r(A)$. If a matrix $B$ has more indices vanishing, more than $r(A)$ columns will be complete zero.
If $A$ is nilpotent, then the total number of nonvanishing indices of cardinal form is no more than $n-1$. Assume $B$ has less indices without vanishing, we claim that by change the order of rows and columns, $B$ is in the form of $\left(\begin{matrix}B_1 & \\ & B_2\end{matrix}\right)$, and then we can use induction on $B_1$ and $B_2$. Now, we are going to illustrate this claim. Otherwise, we can find some indices $(1,i)$ or $(i,1)$ with $i\neq 1$ nonvanishing. By change of order, we can assume $i=2$. Then, we can find some indices $(k,j)$ or $(j,k)$ with $k\in \{1,2\}$ and $j\notin \{1,2\}$, and similarly, we can assume $j=3$. Just proceeding this process, finally, we will find that $B$ cannot have nonzero indices less than $n-1$.
It seems to be subtle when dealing with general case. Even for the matrix similar to a single Jordan block, I cannot figure out a proof.
Denote $\# A$ to be the number of nonvanishing indices. A nearly trivial observation may be useful $$\# (A+B)\leq (\# A)+(\#B)\qquad \#(AB)\leq (\# A)\cdot (\# B)$$