Question : For every even $k\ge 4$, is the following $(\star)$ true?
$$\begin{align}\text{If $\beta=0.{a_1}^{k}{a_2}^{k}{a_3}^{k}\cdots\in\mathbb Q$, then $\alpha=0.a_1a_2a_3\cdots\in\mathbb Q$.}\qquad(\star)\end{align}$$
Here, $a_i$ is the $i$-th decimal place of $\alpha$.
Example : For $k=2,\alpha=0.12345,$ we have $\beta=0.1491625.$
Motivation : I've been able to prove that the answer is NO for $k=2$, and that the answer is YES for every odd $k\ge 3$. The answer seems YES for any $k\ge 3$, but I'm facing difficulty for treating every even $k$ in general. Can anyone help?
Update : I crossposted to MO.
P.S. The followings are the counterexamples for $k=2$. $$\alpha=0.2372377237772377772377777\cdots\Rightarrow \beta=0.\overline{49}.$$ $$\alpha=0.1842184242184242421842424242\cdots\Rightarrow \beta=0.\overline{164}.$$ $$\alpha=0.183423471831834234718318318342347\cdots\Rightarrow \beta=0.\overline{1649}.$$