Conjugation map in quadratic integer rings is an automorphism

Question

Let $d$ be a square free integer such that $d \equiv 1 \mod 4$ and let $R$ be the ring $$ R = \left\{a + b\frac{1+\sqrt{d}}{2} \mid a, b \in \mathbb{Z}\right\} $$ I am trying to show that the map \begin{align*} \phi:& R \longrightarrow R\\ a + b\frac{1+\sqrt{d}}{2} &\mapsto a + b\frac{1-\sqrt{d}}{2} \end{align*} is an automorphism of $R$. This would then imply that there exists an integer $c$ such that $$ b\frac{1-\sqrt{d}}{2} = c\frac{1+\sqrt{d}}{2} $$ What can $c$ be?