The notation I am using is:
$S_{4}$: the permutation group of order 4
$\mathrm{Aut}(U_{8})$: the set of all automorphisms on the set $U_{8}$
$U_{8}$: the group of numbers relatively prime to 8
I know that $U_{8} = {{1,3,5,7}}$.
Forgive my lack of a proper table, but the "multiplication table" for $U_{8}$ is given by:
$\begin{bmatrix} 1*1 & 1*3 & 1*5 & 1*7 \\[0.3em] 3*1 & 3*3 & 3*5 & 3*7 \\[0.3em] 5*1 & 5*3 & 5*5 & 5*7 \\[0.3em] 7*1 & 7*3 & 7*5 & 7*7 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 5 & 7 \\[0.3em] 3 & 1 & 7 & 5 \\[0.3em] 5 & 7 & 1 & 3 \\[0.3em] 7 & 5 & 3 & 1 \end{bmatrix}$
So I believe I should begin by asking how many elements does $\mathrm{Aut}(U_{8})\ $ have. An element in $\mathrm{Aut}(U_{8})\ $ is an isomorphism from $U_{8}$ to itself. My research led to some discussions of finding the number of generators of the group - however $U_{8}$ is not cyclic so this falls apart for me.
Can somebody point me in the right direction? Even just explaining the different possibilities would help me tremendously. Other than the trivial automorphism $\Phi(x)=x\ $ I cannot think of anything.
I should also note that $U_{8}$ is isomorphic to the Klein four-group. I'm not sure how this helps me.
Let's look at the automorphisms of $V$, the Klein vierergruppe in the form:
$V = \{e,a,b,c\}$ where $a^2 = b^2 = c^2 = e$, and $ab = ba = c$.
Note that any such automorphism must permute the set $\{a,b,c\}$ since an element of order $2$ has to go to an element of order $2$ under such an automorphism.
This means $\text{Aut}(V)$ is isomorphic to a subgroup of $S_3$, which can be embedded (in several different ways) in $S_4$.
Another way to realize $\text{Aut}(V)$ as a subgroup of $S_4$ is to regard any such automorphism as a permutation (since automorphisms are bijective) of the elements of $V$, which has $4$ elements.