$f(x)=x^3+x^2+1$ over $Z_2[x]$ what are the automorphisms of this field???
what is known the field $F=Z_2[x]/(f(x))$ has order of $|F|=3^2=8$
letting $a$ be a zero rot of $f(x)$ in F
$$ \begin{aligned} F&=\{0,1,a,a^2,a^3,a^4,a^5,a^6\} \\&=\{0,1,a,a+1,a^2,a^2+a+1,a^2+1,a^2+a\} \end{aligned}$$
if $a$ is a zero then $a^2,a^4$ are also zeros
so $$f(x)=(x-a)(x-a^2)(x-a^4)$$
Theorem1
$$[GF(p^n):GF(p)]=n $$
th2 For each divisor $m$ of $n$, $FF(p^n)$ has a unique subfield of order $p^m$, moreover these are the only subfields of $GF(p^n)$
I am not sure that helps for the question being find all the automorphisms for the field but it's something
Attempt1
so it would be something like $$ Z(a,a^2,a^4)$$
Taking a wild guess its $8$ the lcm??
Attempt 2
Möbius inversion formula
$p=3,n=2$ the divisors of $p$ are 3 and 1
$$ \psi (2)=\frac{1}{3}[\mu(1) (3)^3+ \mu(3) (3)^1] = \frac{1}{3}[ (3)^3- (3)^1]=27-3=24$$
Hint:
Your attempt 1 is the right track. Consider the Frobenius homomorphism of the field $F$, $\operatorname{Fr}\colon x\longmapsto x^2$. Iterating this homomorphism, you obtaid the $3$ roots of the polynomial $x^3+x+1$, and as $F$ is a simple extension of $\mathbf F_2$, this means the polynomial splits in $F$.
Therefore, since automorphisms of a simple extension are entirely determined by the image of its primitive element, its Galois group is generated by $\operatorname{Fr}$. You can easily check that $\operatorname{Fr}^3=\operatorname{Id_F}$, so the Galois group is isomorphic to $\mathbf Z/3\mathbf Z$.