Using some suggestions from the other commenters:
The alternating group, $A_4$, has the set $H=\{I,(12)(34),(13)(24),(14)(23)\}\cong V_4$ as a subgroup. If $f\in S_4\supseteq A_4$ is a permutation, then $f^{-1}[(12)(34)]f$ has the effect of swapping $f(1)$ with $f(2)$ and $f(3)$ with $f(4)$. One of these is $1$, and depending on which it is paired with, the conjugated element may be any of $H-\{I\}$, since the other two are also swapped. Thus $H\lhd S_4$ is normal, so $H\lhd A_4$ as well. Similarly, $H$ has three nontrivial subgroups, and taking $K=\{I,(12)(34)\}\cong C_2$, this is normal because $V_4$ is abelian. But $K\not\lhd A_4$, since $$[(123)][(12)(34)][(132)]=(13)(24)\in H-K.$$
Moreover, this is a minimal counterexample, since $|A_4|=12=2\cdot 2\cdot 3$ is the next smallest number which factors into three integers, which is required for $K\lhd H\lhd G$ but $\{I\}\subset K\subset H\subset G$ so that $[G\,:\,H]>1$, $[H\,:\,K]>1$, $|K|>1$ and
$$|G|=[G\,:\,H]\cdot[H\,:\,K]\cdot|K|.$$
The smallest integer satisfying this requirement is 8, but the only non-abelian groups with $|G|=8$ are the dihedral group $D_4$ and the quaternion group $Q_8$, and neither of these have counterexamples. (Note that if $G$ is abelian, then all subgroups are normal.) Thus $A_4$ is a minimal counterexample. (Edit: Oops, $D_4$ has a counterexample, as mentioned in the comments: $\langle s\rangle\lhd\langle r^2,s\rangle\lhd \langle r,s\rangle=D_4$, but $\langle s\rangle\not\lhd D_4$.)
However, if $H\lhd G$ and $K$ is a characteristic subgroup of $H$, then $K$ is normal in $G$. This is because the group action $f$ defines an automorphism on $G$, $\varphi(g)=f^{-1}gf$, and because $H$ is normal, $\varphi(H)=H$ so that $\varphi|_H$ is an automorphism on $H$. Thus $\varphi(K)=K$ since $K$ is characteristic on $H$ and so $\{f^{-1}kf\mid k\in K\}=K\Rightarrow K\lhd G$.
You are on the right track. Try to be a bit more straightforward in your answer. For instance, assuming that $H$ is normal in $G$ and then prove directly that $H$ is a union of conjugacy classes. To achieve that, you need to find the conjugacy classes that make up $H$. Consider, just one element $h\in H$. Since $H$ is normal, you know that $g^{-1}hg\in H$ for all $g\in G$. Doesn't that mean the $CC(h)$ is contained in $H$? So, you found a conjugacy class that is fully contains in $H$ and it contains the element $h$. But $h$ was arbitrary, so it is true in general that if $h\in H$ then its conjugacy class $CC(h)$ satisfies $h\in CC(h)\subseteq H$. Now, what does that tell you about $$\bigcup _{h\in H}CC(h)=?$$
Formalize carefully the arguments above and you'll have a proof of the first direction. You can then finalize the proof in the other direction.
Best Answer
Using @user29743 hint, let $a = bh$. Then $h = b^{-1}a \in H$ means $ba^{-1} = bh^{-1}b^{-1} \in H$, and since $H$ is a subgroup, $(bh^{-1}b^{-1})^{-1} = bhb^{-1} \in H$. Replacing $b$ as $g$ we have: $ ghg^{-1} \in H$.
It's actually the case that $gHg^{-1} = H$ for every $g$, since if $gHg^{-1} \subset H$, by letting $y = g^{-1}$, we get $yHy^{-1} \subset H \subset y^{-1}Hy =gHg^{-1}$.