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. 



Changing values of digits of an algebraic irrational number
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.


