Prove that the group $\operatorname{Aut}(\mathbb{F}_{p^n}^*)$ has order $\varphi(p^n-1)$. We denote $\mathbb{F}_{p^n}^*$ the multiplicative group of the finite field $\mathbb{F}_{p^n}$ and $\varphi$ Euler's arithmotheoretic function.
We know that $\mathbb{F}_{p^n}$ is an extension of the field $\mathbb{F}_p$ with $p^n$ cardinality and the extension is Galois and $\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)= \langle \sigma \rangle$ where $\sigma$ is the Frobenius automorphism.
Also the multiplicative group of $\mathbb{F}_{p^n}$ is cyclic.
Can someone help me to combine these facts to solve this problem?
Thank you in advance!
$\mathbb{F}_{q}^*$ is a cyclic group of order $q - 1$ where $q = p^n$. Since any two cyclic groups of equal order are isomorphic, we know that $\mathbb{F}_{q}^* \cong \mathbb{Z}/(q - 1)\mathbb{Z}$.
Now it is not too hard to show that $\text{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong U(n)$ where $U(n)$ is the group of units under multiplication mod $n$, which when coupled with the statement above, establishes the claim. In order to understand why $\text{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong U(n)$, think about the image of a generator of $\mathbb{Z}/n \mathbb{Z}$ under an automorphism...