Automorphism of Z_p(t) and its invariant field

$\mathbf{Problem:}$ Let $E=\mathbb{Z}_p(t)$ where $t$ is transcedental over $\mathbb{Z}_p$. Let $G$ be the group of automorphism generated by the automorphism of $E$ such that $t\to t+1$. Determine $F=\text{Inv}G$ and $[E:F]$.

$\mathbf{Proof:}$ Suppose $\sigma$ is the automorphism of $E$ that $t\to t+1$.  Group $G$ is generated by $\sigma$. It is easy to verify that $G$ is the cyclic group $\{1,\sigma,\cdots,\sigma^{p-1}\}$. So If $F=\text{Inv}G$, then $[E:F]=|G|=p$.

Let’s determine $F$. Note that $\mathbb{Z}_p(u)$ where $u=t^p-t$ is invariant under $\sigma$, because $a^p=a$, for $\forall \,a\in \mathbb{Z}_p$.  So $\mathbb{Z}_p(u)\subset F$.

On the other hand, $\mathbb{Z}_p(u)$ is a field of characteristic $p$ and $E$ is a splitting field of $x^p-x-u$ over $\mathbb{Z}_p(u)$. $x^p-x-u$ has no root in $\mathbb{Z}_p(u)$, thus it is irreducible on $\mathbb{Z}_p(u)[x]$ by a previous conclusion. So $[E:\mathbb{Z}_p(u)]=p=[E:F]$, which means $F=\mathbb{Z}_p(u)$ where $u=t^p-t$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Problem:}$ What if $G$ is replaced by the group of automorphism such that $t\to at+b$, $a,b\in\mathbb{Z}_p$, $a\neq 0$

If $a=1$, $b=0$, then $F=E$ and $[E:F]=1$

If $a=1$, $b\neq 0$, then the conclusion is the same with previous problem.

If $a\neq 1,0$, $|G|=p-1=[E:F]$, $F=\mathbb{Z}_p(u)$ where $\displaystyle u=\prod\limits_{i=1}^{p-1}\sigma^{i}(t)$.

$\mathbf{Remark:}$ Jacobson, Algebra I p243.