Can a matrix have a null space that is equal to its column space?

The nullspace lies inside the domain, while the column space lies inside the codomain. Therefore, if the nullspace is equal to the column space, you must have $m=n$. Also, by the rank-nullity theorem, $n$ must be an even number. It follows that if $n=2k$, the nullspace must be $k$-dimensional. Denote by $\{e_1,\ldots,e_n\}$ the canonical basis of $\mathbb{F}^n$. By a change of basis, we may assume that the nullspace is spanned by $e_1, \ldots, e_k$. Therefore, if the nullspace and column space of $A$ coincide, $A$ must be similar to a matrix of the form $$ A=\pmatrix{0&B_{k\times k}\\ 0&0}, $$ where $B$ is invertible. For instance, consider $A=\pmatrix{0&1\\ 0&0}$ when $n=2$.

$$\pmatrix{2&4\cr-1&-2\cr}$$

Let us study this question in a general setting of $n \times n$ matrices.
For a given $n \times n$ matrix $A$, we denote its nullspace by $\mathcal{N}(A)$, and its column space by $\mathcal{C}(A)$.

Recall that the orthogonal complement of a vector subspace $V$ is $$V^\perp := \{\vec{x} : \forall \vec{v} \in V.~(\vec{x}^T \vec{v} = 0) \}.$$ It is well known that for any $n\times n$ matrix $A$ it holds that $(\mathcal{N}(A))^\perp = \mathcal{R}(A)$, where $\mathcal{R}(A)$ is the row space of $A$. In particular, $\mathcal{C}(A) = \mathcal{R}(A^T)$ implies that $(\mathcal{C}(A))^\perp = \mathcal{N}(A^T)$.

Now let $A$ be any $n \times n$ matrix with $\mathcal{N}(A) = \mathcal{C}(A)$. From our preceding discussion, it is thus necessary and sufficient that $\mathcal{N}(A) = \mathcal{N}(A^T)$. We shall first show that such an $n \times n$ matrix $A$ must be of the form $$Q \left(\begin{array}{cc} X & 0 \\ 0 & 0 \end{array}\right) Q^T,$$ where $X$ is an $r \times r$ matrix whose rank, $r$, is equal to that of $A$, and some bordering $0$'s may be absent if $A$ is of full rank $n$.

$Proof$. Extend any given basis $\{e_{r+1},\ldots,e_n\}$ for $\mathcal{N}(A)$ to a basis $\{e_1,\ldots,e_r,e_{r+1},\ldots,e_n\}$ for $\mathbb{R}^n$. Let $P = (\vec{e}_i^T)$ and $Q = P^{-1}$. Clearly, both $P$ (as well as $Q$) is invertible since its column vectors $\{e_1,\ldots,e_n\}$ are linearly independent. Since $\mathcal{N}(A) = \mathcal{N}(A^T)$, it follows that $PAP^T = \left(\begin{array}{cc} X & 0 \\ 0 & 0 \end{array}\right),$ where $r(X) = r(PAP^T) = r(A) = r$. Hence $A$ is of the desired form. The proof is thus complete.

Now we establish that any $n \times n$ matrix $A$ of the above form has to satisfy the equation $$\mathcal{N}(A) = \mathcal{N}(A^T),$$ and hence the condition that $\mathcal{N}(A) = \mathcal{R}(A)$.

$Proof.$ Suppose $A = Q \left(\begin{array}{cc} X & 0 \\ 0 & 0 \end{array}\right) Q^T$, where $X$ is an invertible matrix with $r(X) = r(A)$. Writing the matrix $\left(\begin{array}{cc} X & 0 \\ 0 & 0 \end{array}\right)$ as $Y$, we have that $\vec{x} \in \mathcal{N}(A)$ iff $QYQ^T \vec{x} = 0$ iff $YQ^T \vec{x} = 0$. Now since $X$ is an invertible matrix of size $r$, the last condition is equivalent to $Y^TQ^T \vec{x} = 0$, which in turn is equivalent to $QY^TQ^T \vec{x} = 0$ iff $x \in \mathcal{N}(A^T)$. This shows that $\mathcal{N}(A) = \mathcal{N}(A^T)$.