Let $\mathscr{C}_n$ be the lattice of [clones](https://en.wikipedia.org/wiki/Clone_(algebra)) on the $n$-element set $\{1,...,n\}$. $\mathscr{C}_2$ is [complicated but countable](https://en.wikipedia.org/wiki/Post%27s_lattice), but $\mathscr{C}_3$ (and all higher lattices) [is of size continuum](https://math.stackexchange.com/questions/4663231/is-there-a-clone-on-a-three-element-set-which-is-not-finitely-generated/4664400#4664400).

> **Question**: Does $\mathscr{C}_4$ embed as a lattice into $\mathscr{C}_3$?

At [MSE](https://math.stackexchange.com/questions/4718501/lattices-of-clones-is-4-worse-than-3?noredirect=1&lq=1) I asked a harder version of this question, but per Keith Kearnes' answer there it seems that even this weaker question may be tricky (or even open).