Q:

algebraic conjugate

Let $\alpha, \beta$ be real roots of an irreducible polynomial over the field of rational numbers (i.e., $\alpha, \beta$ are algebraic conjugates). 
Is it possible that $\beta=\alpha^2$?

A:

The answer is no, unless $\beta = \alpha \in \{0,1\}$.
Let $K$ be the splitting field of the irreducible polynomial. There is some automorphism $\phi$ of $K$ such that $\phi(\alpha) = \beta = \alpha^2$. Then we also have $\phi\left(\alpha^m\right) = \phi(\alpha)^{m} = \alpha^{2m}$. By induction, $\alpha^{2^n}$ is a root of the polynomial for all $n\geq 0$. Thus, since $\alpha$ is real, the polynomial has infinitely many roots, a contradiction, except if $\alpha \in \{-1,0,1\}$. The case $\alpha = -1$ cannot occur.
(If complex values are allowed, the roots of $x^2 + x + 1$ give a counterexample.)

