Suppose $\theta$ is a root of the polynomial $x^{10}-2$, then determine the Automorphism group of $K=\mathbb{Q}(\theta)$.
What I am thinking is $\mathbb{Q}(\theta)$ automorphism is uniquely determined by image of $\theta$. Any root of $x^{10}-2$ is of the form $2^{\frac{1}{10}}\zeta^k$ where $\zeta$ is $10$th primitive root of unity.
Also since $2^{\frac{1}{10}}\zeta$ is a root, then so is $-2^{\frac{1}{10}}\zeta$ and this is of the form $2^{\frac{1}{10}}\zeta^k$. But I am unable to determine this value of $k$ and would appreciate if someone can guide.
Also how do I show that identity and the conjugation are the only automorphisms? I think something like $A(2^{\frac{1}{10}}\zeta)=2^{\frac{1}{10}}\zeta^k$ but I cannot conclude anything from here.
Let $\theta=\sqrt[10]{2}\in\mathbb R$.
Let $\sigma\colon\mathbb Q(\theta)\to\mathbb Q(\theta)$ be an automorphism, so that $\sigma(\theta)^{10}=\sigma(\theta^{10})=\sigma(2)=2$. Thus, since $\sigma(\theta)\in\mathbb Q(\theta)\subset\mathbb R$, we conclude $\sigma(\theta)=\pm\theta$. Moreover, an automorphism $\sigma$ is uniquely determined by the value of $\sigma(\theta)$.
Thus, the automorphism group has order $2$.