Restrict attention to **Z**_{p},
and remember that its multiplicative group has a generator.
Let r be such a generator.
If we raise r to the p-1 power we get 1.
We must raise r to the p to get r back again.
Raising r to a different power gives another integer, but all integers must map to themselves.
That means every automorphism raises all elements to the j power,
where j is some multiple of p.

The obvious choice is the frobenius automorphism, which raises everything to the p power. This is indeed a field automorphism, and by the above paragraph, it generates all possible automorphisms.

If c() is the frobenius automorphism,
it can be invoked again and again,
until we reach c^{n},
which raises everything to the p^{n}.
Remember that the powers of b cover the nonzero elements of F,
all p^{n}-1 of them,
so when we raise b to the p^{n}, we get b back again,
giving the trivial automorphism.
The group of automorphisms is a cycle of length n.

If K is a subfield of F it is finite,
with order p^{r} for some r.
If K and L are two subfields of order p^{r},
with K ≠ L,
then there are more than p^{r} elements satisfying
x^{(pr)} = x, which is impossible.
There is at most one subfield of order p^{r}.
Remember that p^{r} - 1 must divide p^{n} - 1, hence r must be a factor of n.
You can show this by synthetic division.
To illustrate, divide p^{2}+p+1 (which is p cubed minus 1) into p^{4}+p^{3}+p^{2}+p+1 (which is p to the fifth minus 1).
The first term of the quotient is p^{2}, giving a remainder of p+1, which is smaller than the divisor.
It doesn't work because 3 does not divide 5.
Thus r must divide n.

For any r dividing n,
let d = c^{r}.
Thus d is the automorphism produced by raising everything to the p^{r}.
The field fixed by this automorphism
is precisely the elements satisfying
x^{(pr)} = x.
Aside from 0,
the roots of this equation are generated by
b^{(pn-r)}.
this establishes the subfield of order p^{r}.

The subfields of F are precisely the fields fixed by the various subgroups of the group of automorphisms on F; one subfield and subgroup for each r dividing n.

If K is an intermediate extension of order p^{s},
the field extensions between K and F
are the subfields of F that contain all of K.
These are the subfields of order p^{r},
where s divides r and r divides n.
Once again these subfields are fixed by the subgroups
of the group of automorphisms of F that fix K.

All this may seem confusing for now, but once you are familiar with Galois extensions, it will seem straightforward. We are really saying every finite extension of a finite field is Galois.

Let's use this example to dispel a common myth.
In a nice galois extension, the conjugates need not span the extension as a vector space.
Adjoin u^{2}+1 to **Z**_{3}.
Since 2 is not a square, this is irreducible.
The sum of the roots is given by the second coefficient, which is 0.
Thus u+u^{3} = 0.
The conjugates are not linearly independent, and cannot span the entire vector space of dimension 2.

Let's explode another myth - the adjoined element need not be primitive.
In the above example, u^{4} = 1.
Thus u does not have order 8.
A primitive element might be u+2, whose square is u.