TITLE: Uniqueness of characteristic polynomial of linear transformation in finite fields
QUESTION [5 upvotes]: Let $T : V \to V$ be a linear transformation of the $F$-vector space $V$. Then using the (abstract) determinant function $\det : \operatorname{Hom}(V, V) \to K$ we can define a function
$$
  \lambda \mapsto \det(\lambda \cdot \operatorname{id} - T)
$$
from $F$ to $F$. Now if we represent $T$ by a matrix $A$ we then have $\det(\lambda \cdot \operatorname{id} - T) = \det(\lambda \cdot I - A)$ where on the RHS the determinant function is an expression in the entries of the matrix. Now if we change the basis, i.e. represent $T$ by a different matrix $B = S^{-1}AS$, then a simple calculation shows that
$\det(\lambda \cdot I - B) = \det(S^{-1}(\lambda I - A)S) = \det(\lambda \cdot I - A)$ and hence the value of of the matrix expression does not depend on the basis.
But this is often used as a justification that the characteristic polynomial (i.e. the polynomial $\det(\lambda \cdot I - A)$) is independent of the of the choosen basis. But all I can derive from the above arguments is that the values of the determinant are the same, i.e. if $p(x) := \det(x\cdot I - A)$
and $q(x) := \det(x \cdot I - A)$, then $q(x) = p(x)$ for all $x \in F$ if $B = S^{-1}AS$. If $F$ is infinite, this gives that the polynomials are equal (i.e. have the same sequence of coefficients, and hence the coefficients represent also invariants of the transformation). 
But the equality that for $p(x) = a_n x^n + \ldots + a_1 x + a_0$, $q(x) = b_m x^m + \ldots + b_1 x + b_0$ we have
$$
 p(x) = q(x) \quad \mbox{ iff } \quad m = n, ~ a_i = b_i, ~ i = 0,\ldots, n
$$
does not need to hold in finite fields, for example $p(x) = x^2 + x$ and $q(x) = 0$ in $\mathbb Z/2\mathbb Z$.
So then, is the characteristic polynomial (as a formal polynomial, i.e. determined by its coefficients) still unique in the case of finite fields? And if not, do you know an example?

REPLY [2 votes]: But all I can derive from the above arguments is that the values of the determinant are the same, i.e. if $p(x) := \det(x\cdot I - A)$
  and $q(x) := \det(x \cdot I - A)$, then $q(x) = p(x)$ for all $x \in F$ if $B = S^{-1}AS$.

There's your mistake: you want to do this calculation for the specific element $x \in F[x]$, or if you insist on working only over fields, for the element $x \in F(x)$.