What are some relationships between a matrix and its transpose? All I can think of are that
If symmetric, they're equivalent
If A is orthogonal, then its transpose is equivalent to its inverse.
They have the same rank and determinant.
Is there any relationship between their images/kernels or even eigenvalues?

There is a very strong connection between an $m\times n$ matrix $A$ and its transpose $A^T$, at least if the base field is $\def\R{\mathbb{R}}\R$.
Denote by $C(A)$ the column space, that is, the subspace of $\R^m$ generated by the columns of $A$. Denote by $N(A)$ the null space, that is, the set of vectors $v\in\R^n$ such that $Av=0$.
Associated to a matrix $A$ we have four subspaces:


*

*$C(A)\subseteq\R^m$

*$N(A)\subseteq\R^n$

*$C(A^T)\subseteq\R^n$

*$N(A^T)\subseteq\R^m$


Now the relations (of course, the second one is equivalent to the first)
$$
C(A)\oplus N(A^T)=\R^m,\qquad
C(A^T)\oplus N(A)=\R^n
$$
which means that every vector of $\R^m$ can be written in one and only one way as the sum of a vector in $C(A)$ and one in $N(A^T)$.
To see why, consider a vector $v\in C(A)\cap N(A^T)$; then $v=Ax$ for some $x\in\R^n$ because $v\in C(A)$; since also $v\in N(A^T)$ we get
$$
0=A^Tv=A^TAx
$$
from which we deduce
$$
0=x^T0=A^TAx=(Ax)^T(Ax),
$$
so that $Ax=0$ and therefore $v=Ax=0$. Thus $C(A)\cap N(A^T)=\{0\}$ and we only need to show that $C(A)+N(A^T)=\R^m$.
The relation between the rank of $A$ and of $A^T$ and the "rank-nullity theorem" tell then that
$$
\dim N(A^T)=m-\mathrm{rk}\,A^T=m-\mathrm{rk}\,A=m-\dim C(A),
$$
so that $\dim C(A)+\dim N(A^T)=m=\dim\R^m$ and hence $C(A)+N(A^T)=\R^m$.
The same holds for matrices over $\def\C{\mathbb{C}}\C$, but using the Hermitian transpose instead of the transpose.

This doesn't hold however for matrices over an arbitrary field, because the step “$(Ax)^T (Ax)=0$ implies $Ax=0$” cannot be done in general.
