Problem
Let $\beta$ be a non-trivial aoutomorphism of the field $F$. Use $\beta$ to construct an outer automorphism of $GL(n,F)$. Do all outer automorphisms of $GL(n,F)$ arise in this way?
My attempt
What do we know? We know that $\beta$ is a bijective homomorphism and non-trivial.
Observation: Center of $GL(n,F) = \{aI_n\},\ a\in F$. So we need a map that doesn't fix the elements in the center because we need to find an outer automorphism (not inner automorphish)
Let us consider the map
$\phi:\ GL(n,F) \rightarrow GL(n,F) $
$a I_n \mapsto \beta(a) I_n$
We need to show that $\phi$ is an outer automorphism.
To show that $\phi$ is bijective is trivial, it follows from the fact that $\beta $ is bijective.
$\phi$ is a homomorphism:
$\phi(aI_nbI_n)=\phi(abI_n)=\beta(ab)I_n=\beta(a)\beta(b)I_n= \beta(a)I_n\beta(b)I_n = \phi(aI_n)\phi(bI_n)$
And finally $\phi$ is an outer automorphism:
We need to show that it is not inner, that is that it does not fix elements of the Center.
$\phi(aI_n)=\beta(a)I_n\neq a I_n$, since $\beta(a) \neq a\ , \forall \ a \in F$ by the assumption.
Question
Please correct me if I am wrong in some part of the proof and I am stuck in "Do all outer automorphisms of $GL(n,F)$ arise in this way?". My guess is that all the automorphism arise in this way since this the only way I can think of, when constructing function which doesn't fix elements of Center.
The answer to your question is no, because the outer automorphism that maps each matrix $A$ to $(A^{\mathrm T})^{-1}$ (inverse of transpose) does not arise in this way. (This can also be seen as the graph automorphism of the $A_{n-1}$ Coxeter diagram, which is associated with this group.)
This map induces $x \mapsto x^{-1}$ on the centre of the group (the scalar matrices). This shows that it is an outer automorphism whenever $|F| > 2$. In fact it is an inner automorphism on ${\rm GL}(2,2) \cong S_3$, but outer on ${\rm GL}(n,2)$ for $n>2$, because it maps stabilizers of spaces of dimension $k$ to stabilizers of spaces of dimension $n-k$.