Changing values of digits of an algebraic irrational number Consider the set of irrationals $\mathbb{I} \cap (0,1)$. Define the map $f$ that takes
$x= \sum_{n\ge 1} \frac{a_n}{3^n}$ to $\sum_{n\ge 1} \frac{ \phi(a_n)}{3^n}$, where $\phi$ is the permutation of $\{0,1,2\}$, $0\mapsto 1 \mapsto 2\mapsto 0$.  Is it possible that both $x$ and $f(x)$ are algebraic for some $x$ ? 
The common belief is that they are "random". A solution would not run contrary to that. Still my feeling is that this would not be possible. 
 A: Let $x_j = \sum_{n \geq 1} \mathrm{1}_{a_n=j} 3^{-n}$. Then 
$$
x_0 + x_1 + x_2 = \frac{1}{2} \\
x_1 + 2x_2 = x \\
x_0 + 2x_1 = f(x).
$$
If $x$ and $f(x)$ are algebraic then the equations above would yield that $x_j$ is algebraic for each $j$. It is conjectured and generally believed that any irrational algebraic number is normal in all bases. Since $x_j$ is obviously not normal in base $3$, this conjecture would imply that $x_j$ is rational for each $j$, and thus that $x$ is rational.
A: Just to augment @js21's answer. The question of whether irrational $x_i$ are necessarily transcendent is equivalent to this question. As the answer suggests, even this specific question is wide open, or at least was such in 2012.
As for the other bases $b>2$, the result is similar. Defining $x_i$ as in js21's answer, you get that
$$
  x_{b-1}=\frac{x+\frac1{b-1}-f(x)}{b}
$$
is algebraic though not normal. (Surely, if $x_{b-1}$ is just rational, then $x-(b-1)x_{b-1}$ will be an algebraic irrational though not normal.)
