From discussions I had with Sam, Scott, and Jack:

To solve the problem, it would suffice to find a reflexive domain $X$ with a retract onto $[0,1]$.

This is because if you have a reflexive domain $X$, that is, an $X$ with a continuous surjective map $f::X\to X^X$, and $A$ is a retract of $X$, then there's also a continuous surjective map $g::X\to A^X$.

Proof: If $A$ is a retract of $X$ then we have a retraction $r::X\to A$ and a section $s::A\to X$ with $r\circ s=1_A$. Construct $g\left(x\right):=r\circ f\left(x\right)$. To show that $g$ is a surjection consider an arbitrary $q\in A^X$. Thus, $s\circ q::X\to X$. Since $f$ is a surjection there must be some $x$ with $f\left(x\right)=s\circ q$. It follows that $g\left(x\right)=r\circ f\left(x\right)=r\circ s\circ q=q$. Since $q$ was arbitrary, $g$ is also a surjection.

"Self-Reference and Fixed Points: A Discussion and an Extension of Lawvere's Theorem" by Jorge Soto-Andrade and Francisco J. Varela seems like a potentially relevant result. In particular, they prove a converse Lawvere result in the category of posets (though they mention doing this for $[0,1]$ in an unsolved problem.) I'm currently reading through this and related papers with an eye to adapting their construction to $[0,1]$ (I think you can't just use it straight-forwardly because even though you can build a reflexive domain with a retract to an arbitrary poset, the paper uses a different notion of continuity for posets.)