Let $k$ be an algebraically closed field. It is well-known that the isom-sheaf Isom$(\mathbb{A}^1_k,\mathbb{A}^1_k)$ is not representable by an algebraic space. (To be clear, the functor Isom$(\mathbb{A}^1,\mathbb{A}^1)$ associates to a $k$-scheme $S$, the set of isomorphisms $\mathbb{A}^1_S\to \mathbb{A}^1_S$ of $S$-schemes.)

Now, I expect that the non-representability of this isom-functor implies the non-representability of many hom-functors. My question is about how to make this precise.

Let X be a (positive-dimensional) variety and let $f:\mathbb{A}^1_k\to X$ be a finite morphism. How does one show that the hom-functor $\mathrm{Hom}_k(\mathbb{A}^1_k,X)$ is not representable by an algebraic space?

What have I tried? Well, there is a natural morphism Isom$(\mathbb{A}^1,\mathbb{A}^1) \to \mathrm{Hom}(\mathbb{A}^1_k,X)$ which sends $g$ to $f\circ g$. My expectation is that this is an open or closed immersion of functors. But how to make this precise?