A different approach, by the principle of irrelevance of algebraic inequalities:

a) **Over an infinite field**:

1) **Diagonalizable matrices:** If $A$ is diagonalizable to $D$ via $P$ then it is similar to its transpose, since $D$ being symmetric implies
$PAP^{-1}=D=D^T=(PAP^{-1})^T = P^{-T}A^TP^T,$
so that
$$A^T=QAQ^{-1}, Q:=P^TP.$$

2) **Similarity is polynomial:** Two matrices A,B are similar iff they have the same invariant factors $\alpha_i(A)=\alpha_i(B)$, which are polynomials in the entries of the matrices.

3) **Extension to all matrices:** Let $f_i$ be the polynomial $\alpha_i(A)-\alpha_i(A^T)$ on the entries of $A$.

Consider the set of matrices with pairwise different eigenvalues, which are diagonalizable. These are precisely those which do not annihilate the discriminant of their characteristic polynomial, which is a polynomial $g$ in the entries of the matrix.

Thus we have that $g(A)\neq0$ implies $f_i(A)=0$. By the irrelevance of algebraic inequalities, $f_i(A)=0$ for all matrices, i.e., $A^T$ is similar to $A$.

b) **Over a finite field:**

Matrices over a finite field $K$ can be seen as matrices over the infinite field $K(\lambda)$ ($\lambda$ trascendental over $K$), so by part a) they also satisfy the result.