This is the problem I'm trying to solve:
Let $K/F$ be a field extension, and let $f: K → K'$ be an isomorphism of $K$ with a field $K'$ which maps $F$ onto the subfield $F'$ of $K'$. Prove that the map $g ↦ fgf^{-1}$ defines a group isomorphism $Aut(K/F) → Aut(K'/F')$. ($Aut(K/F)$ denotes the group of field isomorphisms $g: K → K$ which restrict to the identity map on $F$.)
I'm having trouble figuring out how to begin proving this. I sort of understand what it's asking for, but I don't really understand the structure of $Aut(K/F)$, and somehow trying to think about what's actually happening in this problem confuses me.
Task. This task doesn't actually require you to know alot about the theory of field extensions, although you will need some familiarity with the terms group and field, and homomorphisms and isomorphisms of both. To get into this task, first observe that $Aut(K/F)$ as defined in your question is indeed a group with the usual composition of functions as group operation. Afterwards show that for a $g\in Aut(K/F)$ (g:K\to K) the term $fgf^{-1}$ yields indeed an element of $Aut(K'/F')$, i.e. that $fgf^{-1} : K' \to K'$ is a field isomorphism which restricts to identity on $F'$. To complete the actual task, define $\varphi:Aut(K/F) \to Aut(K'/F')$ by $\varphi(g) = fgf^{-1}$. Then $\varphi$ is a map between groups and you are to show that this map is a group isomorphism, so show that the map is a homomorphism and a bijection. I'll explain this in detail later, but you should try to verify all this by yourself.
Example. For a simple example of such a group consider $Aut(\mathbb C / \mathbb R)$. This consists of bijective maps of $\mathbb C$ which leave the real numbers fixed and which preserve addition and multiplication. The identity function and the complex conjugate satisfy these conditions, because $\overline a + \overline b = \overline{a+b}$ and $\overline{a}\overline{b} = \overline{ab}$. You can show that only these homomorphisms satisfy the conditions, because if $h\in Aut(\mathbb C / \mathbb R)$ then $h(i)^2 = h(i\cdot i) = h(-1) = -1$, so $h(i) = i$ or $h(i) = -i$, but then $h(a+ib) = a + h(i) b = a + ib$ ($h=id$) or $h(a+ib) = a + (-i)b$ which yields the complex conjugate. so $Aut(\mathbb C / \mathbb R)$ is a two-element-group.
I can't seem to come up with an elementary application of the theorem right now. If you're desperate for examples you could try to prove this for noncommutative fields/division rings (not sure if it holds) and find a third automorphism of the quaternions that leaves the reals fixed. Then the galois groups of the two extensions are different and by the contraposition of the theorem there's no ring isomorphism between the complex numbers and the quaternions that leaves the reals fixed, which admittedly isn't that surprising.
Proof. Generally the isomorphisms which map from $K$ to $K$ form a group, so we are only to show that Aut(K/F) forms a subgroup of that group. If $f, g \in Aut(K/F)$, then $f(x) = x$ and $g(x) = x$ for $x\in F$, so $gf(x) = x$ for $x\in F$, thus $gf \in Aut(K/f)$, so it is closed under composition. The identity is a field isomorphism which fixes $F$, because it fixes $K$. Finally the inverse of an isomorphism $f \in Aut(K/F)$ will fix $x$, because if $x \in F$ then $f(x) = x$ and for the inverse $f^{-1}$ of $f$ the identity $f^{-1}f = id$, thus $f^{-1}f(x) = x$ holds, therefore $f^{-1}f(x) = f^{-1}(x) = x$ for $x \in F$.
Now for the actual task. We are given an Isomorphism $f: K \to K'$ with $f(F) = F'$ and we define $\varphi:Aut(K/F) \to Aut(K'/F')$ by $\varphi(g) = fgf^{-1}$. We've seen that $Aut(K/F)$ and $Aut(K'/F')$ are groups, now show that $\varphi$ is an isomorphism between those groups. First show that*strong text* $\varphi(g) \in Aut(K'/F')$, i. e. that the statement $\varphi:Aut(K/F) \to Aut(K'/F')$ is actually correct. $\varphi(g) \in Aut(K'/ F')$ is equivalent to $fgf^{-1}(x) = x$ for all $x \in F'$. Now $f$ is a bijection of $F$ and $F'$, so there is some $y \in F$ with $y = f^{-1}(x)$. Then $g(y) = y$ because $g$ restricts to identity in $F$. Thus $fgf^{-1}(x) = fg(y) = f(y) = x$. Thus $\varphi$ is defined correctly.
$\varphi$ is a homomorphism of groups: $\varphi(g)\varphi(g') = fgf^{-1}fg'f^{-1} = fgg'f^{-1} = \varphi(gg')$. $\varphi$ is injective: $\varphi(g) = \varphi(g') \implies fgf^{-1} = fg'f^{-1}$. Compose with $f^{-1}$ on the left and with $f$ on the right side to get $g=g'$. $\varphi$ is surjective: For $h \in Aut(K', F')$, $f^{-1}hf$ lies in $Aut(K, F)$ and $\varphi(f^{-1}hf) = h$. Thus $\varphi$ is an isomorphism of groups.
For further reading look up Galois Theory and Galois Groups.