Topological universality for Cantor maps I am afraid this question might be very naïve, but I find it hard to locate a reference that does not answer a slightly different question.
Consider the Cantor set $C$ and a continous map $f: C\to C$ with finite topological entropy. Is it true that there must exist a $k$ such that the one-sided full shift on an alphabet of $k$ letters $\sigma : \{1,\dots, k\}^{\mathbb{N}} \to \{1,\dots, k\}^{\mathbb{N}}$ contains an embedded copy of $f$?
 A: No.
Your condition is called being a (topological) subshift. If $(C,f)$ is a topological subshift, then there exists a finite clopen partition $P$ of $C$ such that the family $(f^{-n}P)_{n\ge 0}$ separates the points of $C$. (Indeed this is obvious of the shift on $k$ letters: take the partition into $k$ clopen subsets defined by evaluation at 0).
An example with entropy zero that is not a topological subshift is the odometer $(\mathbf{Z}_p,+1)$. Indeed, at the opposite, it satisfies that for every $P$, $(f^{-n}P)_{n\ge 0}$ can be refined to a finite partition (classes modulo $p^n$ for some $n$, since the latter partition is $f$-invariant). Also this shows that its entropy is (obviously) zero.
That $(C,f)$ is a topological subshift is also referred as "$f$ is expansive", but I find this terminology counter-intuitive (at least to me). 

Algebraic insight: A pair $(C,f)$ is the same, through Stone duality, as a pair $(A,f')$ consisting of a Boolean algebra endowed with an endomorphism. The free objects on $k$ generators in this category (Boolean algebra endowed with an endomorphism) correspond to one-sided full shift on $k$ letters. Being a topological subshift thus corresponds to being a quotient of such a free (Boolean algebra with one endomorphism) on finitely many generators. The odometer is a typical case where the corresponding (Boolean algebra with one endomorphism) is not finitely generated; it is indeed infinite and locally finite (increasing union of finite Boolean subalgebras that are stable under the endomorphism).
