Is the following statement always true?
Let $K$ be a finite field, and, for an $n\times n$ matrix $A$ over $K$, \begin{align} \operatorname{tr} A &:= \sum_{i\in[1,n]}^{}{A_{i,i}} \\ \chi_A &:= \text{characteristic polynomial of $A$} \end{align} If $\det A = \det B$ and $\operatorname{tr} A = \operatorname{tr} B$, then $\chi_A = \chi_B$.
Question: I think the statement is wrong but it is intuition. What way is there to find out the answer and to prove it correctly?
Since in two dimensions $\chi_A(t) = t^2-t\operatorname{tr}{A}+\det{A}$, there is no two-dimensional example. In three dimensions, $$ \chi_A(t) = -t^3+t^2\operatorname{tr}{A}-t\frac{1}{2}((\operatorname{tr}{A})^2-\operatorname{tr}{(A^2)})+\det{A}, $$ (and similarly in higher dimensions) so all we have to do is come up with two matrices so that their traces and determinants are the same, but the traces of their squares differ. We may take, for example, the zero matrix and $$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}. $$