Finding a subgroup of the Galois group of $E/(\mathbb{Z}/(p))$ where $E = (\mathbb{Z}/(p))(t)$, $t$ transcendental.

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As the title states, the setup is

Let $E = (\mathbb{Z}/(p))(t)$, and we are looking at it over $\mathbb{Z}/(p)$ and $t$ is transcendental.

Let $G$ be a group of automorphisms of $E$ generated by $\sigma: t \to t+1$. Determine $F = Inv \, G$, and $[E:F]$.

Adjoining a transcendental element to a finite field is one of my weakest aspects of Galois theory, so I don't really know where to start. So $\sigma^p = id$, so $|G| = p$?

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As you have noticed, $\sigma^p=\operatorname{id}$, and clearly $\sigma$ is not identity, so $G$ has order $p$.

To calculate $[E:F]$ recall that the extension degree from the fixed field of the Galois group is always equal to the order of the group.

Finding $F$ shouldn't be too hard. Can you find a single nonconstant polynomial $P(t)$ such that $\sigma(P(t))=P(t)$?