Consider $\mathbb{F}_q$ and the field automorphism $\sigma: \mathbb{F}_q \to \mathbb{F}_q: x \mapsto x^{\sqrt{q}}$.
Suppose I have an element $\lambda \in \mathbb{F}_q$ such that $\lambda = \lambda ^{\sigma}$. This means that $\lambda = \lambda ^{\sqrt{q}} \in \mathbb{F}_{\sqrt{q}}$.
We now that the multiplicative group of $\mathbb{F}_{q}$ is isomorphic to the cyclic group $C_{q-1}$.
Now I want to find an element $\mu \in \mathbb{F}_q$ that is in a way related to $\lambda$. My book states that $\lambda = \mu^{\sqrt{q}+1}$, but I don't see where this expression comes from.
Thanks.
I assume that in your question $q$ is a perfect square and let $\mu$ is a generator for $F_q^* \cong C_{q-1}$. Suppose that $\lambda= \lambda^{\sqrt{q}}$ and let $a < q-1$ be such that $\lambda=\mu^{a}$. Thus we need to solve:
$$ \mu^{a}=\mu^{a \sqrt{q}} \iff \mu^{a(\sqrt{q}-1)}=1.$$
Thus we get $(\sqrt{q}-1)(\sqrt{q}+1)=q-1|(a(\sqrt{q}-1)$ or equivalently, $\sqrt{q}+1|a$.