This answer is really just an expansion of Mohan's and Zach Teitler's comments. I have only provided more detail, that's all. I don't care much for the points and if you want, just give the points to Mohan and Zach Teitler, but I thought it may be useful to the OP and to other readers to provide a complete answer.
Let $\iota: Gr_2(\mathbb{C}^n) \to \mathbb{P}^{d-1}$ be the Plücker embedding, and let
$$f: X \to Gr_2(\mathbb{C}^n)$$
be the map which sends $A \in X$ to the span of its columns.
Then, it can be seen that $\psi = \iota \circ f$. In particular, the image of $\psi$ is the image of the Plücker embedding, which is described by the classical Plücker relations.
If $p \in \operatorname{Im}(\psi)$, how can we describe $\psi^{-1}(p)$? First note that $g \in GL(2,\mathbb{C})$ acts on $X$ by mapping $A$ to $Ag^{-1}$, preserving the fibers of $\psi$. Moreover, since $\iota$ is an embedding, then $\psi^{-1}(p) = f^{-1}(V)$, where $V \in Gr_2(\mathbb{C}^n)$ is the unique point in the Grassmannian which maps to $p$ via the Plücker embedding $\iota$. And $f^{-1}(V)$ is the set of all bases of $V$, which is parametrized by $GL(2,\mathbb{C})$. Thus $\psi^{-1}(p)$ is a copy of $GL(2,\mathbb{C})$.