Suppose $x$ is a word over the alphabet $\{0,1\}$. Let $a$, $b$ be elements of the group Dih$_k$ for some $k$.
Let $\varphi=\varphi_{a,b,k}$ be the map from words over $\{0,1\}$ to elements of the dihedral group Dih$_k$ (having $2k$ elements) such that $\varphi(0)=a$, $\varphi(1)=b$, and $\varphi$ takes concatenation to multiplication: $\varphi(xy)=\varphi(x)\varphi(y)$.
Let's say that $x$ is dihedrally simple if there is some $\varphi_{a,b,k}$ such that $\varphi(x)\ne\varphi(y)$ for all $y\ne x$ of the same length as $x$.
Computer experimentation suggests the
Conjecture: The dihedrally simple words form a regular language, namely $S\cup T$ where $$ S=\bigcup_{n=0}^\infty \{0^n,\quad 0^{n-1}1,\quad (01)^{n/2},\quad 01^{n-1},\quad 01^{n-2}0\} $$ where $(01)^{t+\frac12}=(01)^t0$, and $T$ is obtained from $S$ by interchanging 0 and 1.
My question is: Is the Conjecture similar to anything in the literature? Or do you see how to prove it?
Edit: to answer @LucGuyot's question below: this definition arises in my model of quantum security from a recent [UCNC'17 paper][1] (see also arXiv version) except that there is an additional constraint there, that we map the locked state $|0\rangle$ to the unlocked state $|1\rangle$.
The relevance of dihedral groups is that Dih$_k$ is representable as a subgroup of the projective unitary group $\mathrm{PU}(2, \mathbb C)$. The interpretation then is that $x$ is a secret code which should be punched into a quantum device with buttons labeled 0 and 1 and which trigger unitary operations $U_A$, $U_B$. Any code of the same length as $x$ but different from $x$ will not have the same effect (say, unlocking the device). Any attempt to inspect the state of the device during entering of the code, as one might do with a simple padlock, will constitute a measurement of the device and therefore reset the quantum superposition. [1]: https://link.springer.com/chapter/10.1007/978-3-319-58187-3_12