Let $M_n$ be the algebra of all complex $n \times n$ matrices. The theorem states that
Let $\phi : M_n \to M_n$ be a surjective mapping satisfying $\det(A+\lambda B)=\det(\phi(A)+\lambda\phi(B)),\quad A,B \in M_n, \lambda \in \mathbb{C}.$\
Then there exist $M,N \in M_n$ with $\det(MN)=1$ such that either $\phi(A) = MAN, \; A \in M_n \quad \text{or} \quad \phi(A) = MA^{T}N, \; A \in M_n.$
The proof is pretty long, but I've done proving that $\phi$ is now bijective, rank$(\phi(A))$=rank$(A)$, and the characteristic polynomial of $\phi(A)$ and $A$ are same. But there is another statement says that,
Let us denote by $N ∈ M_n$ the matrix with 1’s above the main diagonal and zeroes elsewhere, $N = E_{12} + E_{23} +···+ E_{n−1,n}$. Since $\phi$ preserves the characteristic polynomial and rank, $\phi(N)$ must be similar to $N$.
The question is, is it necessary for only two conditions to claim that $\phi(N)$ is similar to $N$? Is there any special case for nilpotent matrices?
Here is the reference, here Thanks in advance.