Hint $\ $ Conjugation $\rm \,x\mapsto h(x) = \bar x\,$ is ring homomorphism, i.e. it preserves sums and products $\rm\,\overline{x+y} = \bar x + \bar y,\,\ \overline{xy} = \bar x\,\bar y.\,$ By induction we infer that $\rm\,h\,$ preserves arbitrary compositions of sums and products, i.e. it preserves all polynomial forms. Therefore, if a ring hom $\rm\:h\:$ further fixes the coefficients of a polynomial, then it preserves its roots.
In detail, if $\rm\ h(x\!+\!y) = h(x) + h(y),\ \ h(x\:y) = h(x)\ h(y),\:$ then by induction we deduce
$\rm h(a_0 + a_1 r+\:\cdots + a_n r^n)\ =\ h(a_0) + h(a_1)\ h(r)+\:\cdots+h(a_n)\ h(r)^n =:\, p_{\,h}(h(r))$
therefore $\rm\ h(p(r)) = p_{\,h}(h(r)).\:$ If $\rm\ h\ $ fixes all coefs $\rm\: h(a_{\:k}) = a_{\:k}\:$ then $\rm\ p_{\,h} = p\ $ therefore $\rm\,\color{#c00}{ p(r) =\, 0}\, \Rightarrow\, p_{\,h}(h(r)) = h(\color{#c00}{p(r)}) = h(\color{#c00}0) = 0,\ $ i.e. $\rm\,\ r\:$ root of $\rm\:p\, \Rightarrow\, h(r)\:$ root of $\rm\:p_{\,h}\! = p.$