Being characteristic is transitive Here, I wanted to verify that:

The property of being characteristic is a transitive relation among subgroups of a group $G$.

For subgroups $N,H\leq G$, we have $N$ char $G$ and $H$ char $N$. So $N$ char $G$ implies: $$\forall\psi\in Aut(G); \psi(N)=N$$ $H$ char $N$, so for all elements in $Aut(N)$, $H$ is remained never changing. Especially, when I take $\psi'=\psi|_{N}$ then $\psi':N\to Aut(N)$ would be in $Aut(N)$ and $\psi'(H)=H$. Since the last equality is true for $H$ and the maps, caused by restriction on $N$, then I have $H$ char $G$.
Honestly, I am inly not satisfied form the conclusion here and think I am losing something. Thanks for your hints.
 A: So we have $\,H\,$ char $\,N\,$ char $\,G\,$ . Let 
$$\phi\in \operatorname {Aut}(G)\Longrightarrow \phi(N)=N\,\,,\,\text{since}\,\,N\,\,\text{is characteristic in}\,\,G$$
but this means $\,\left.\phi\right|_N\in\operatorname{Aut}(N)\,$ , so
$$\phi(H)=\left.\phi\right|_N(H)\stackrel{\text{since}\, H\,\mathbf{char}\, N}=H\Longrightarrow H\,\,\mathbf {char}\,G$$
A: I understand your frustration, I also read several proofs but hard to understand them because somehow and I don't know why they all seem to omit the most important part of the explanation.
So we are going to prove that if $H \: char \: N \: char \: G$ then $H \: char \: G$.
Let $\phi \in Aut(G)$, so $\phi(N) = N$ since $N$ is characteristic in $G$.
Now please note the most important part of the proof. We have $\phi|_N \in Aut(N)$ or $\phi|_N(N) = N$, so it's no difference than $\phi(N) = N$ as there is all there the elements of $Aut(N)$ in $Aut(G)$. In other words:

$\phi|_N(N) = \phi(N) = N$.

As $H \: char \: N$, we have the same relationship between $H$ and $N$ as following:

$\phi|_H(H) = \phi|_N(H) = H$.

It is all there elements of $Aut(H)$ in $Aut(N)$, but we have had that $\phi|_N(N) = \phi(N)$, or is all there elements of $Aut(N)$ in $Aut(G)$, including $Aut(H) < Aut(N)$, so it is also all there elements of $Aut(H)$ in $Aut(G)$. In other words we have $\phi|_N(H) = \phi(H)$. Combine them all, we have:

$\phi|_H(H) = \phi|_N(H) = \phi(H) = H$.

Which is by definition $H \: char \: G$.
