Given $N$ char $G$, prove that $N\lt K \lt G$ and $K/N$ char $G/N$ implies $K$ char $G$ 
Let $N$ be a characteristic subgroup of a group $G$. Prove that if $N<K<G$ and $K/N$ is a characteristic subgroup of $G/N$, then $K$ is a characteristic subgroup of $G$.

I tried to prove this as follows:
Since $N$ char $G$ we have
\begin{align}&\alpha(N)=N \text{ for every automorphism } \alpha:G\to G \tag{1}\end{align}
Also $N \vartriangleleft G$
Since $K/N$ char $G/N$ we have
\begin{align} &\beta(K/N)=K/N \text{ for every automorphism }\beta:G/N \to G/N \tag{2}\\
\implies &\beta(yN) \in K/N \text{ for every automorphism } \beta:G/N \to G/N \text{ and } y \in K & \\
\implies & \text{ For every } \beta \in \text{Aut}(G/N),\; y \in K \text{ we have } \beta(yN)= y'N \text{ for some } y' \in K &
\end{align}
Now, let $\phi:G \to G$ be any arbitrary automorphism then we wish to show that $\phi(K)\subseteq K$, i.e., $\phi(x) \in K \; \forall x \in K$
Let $x \in K $ be arbitrary. Now, in order to use $(2)$ or its implications I need to have automorphism of $G/N$. So, if I restrict $\phi$ to the subset $G/N$ of $G$ (wait $G/N$ can not be a subset of $G$. In fact, the cosets of $G/N$ are a subset of $G$ and the factor group $G/N$ is a subset of power set of $G$) so this restriction is absurd. How else can I use $(2)$ then? If I can somehow show $\phi(xN) \in K/N$ then I will be done. Because \begin{align}\phi(xN)&=\{\phi(a) \mid a \in xN\}\\
&=\{\phi(xn) \mid n \in N\}\\
&=\{\phi(x) \phi(n) \mid n \in N\} \quad (\because N,K \subset G \Rightarrow x,n \in G) \\
&=\{\phi(x)a \mid a \in \{\phi(n)| n\in N\}=\phi(N)\}\\
&= \phi(x) \phi(N)\\
&=\phi(x)N \quad (\because \phi(N)=N \text{ from } (1))\end{align}
So, \begin{align} \phi(xN) \in K/N &\implies \phi(x)N \in K/N\\ 
&\implies \phi(x)N=x'N \text{ for some } x' \in K \\
&\implies (\phi(x))^{-1}x' \in N\\
& \implies (\phi(x))^{-1}x' \in K \quad (\because N \subset K)\\
& \implies (\phi(x))^{-1}x'K=K\\
& \implies x'K=(\phi(x))K\\
& \implies K=(\phi(x))K \quad (\because x' \in K)\\
&\implies \phi(x) \in K
\end{align}
Thank You. Any help will be appreciated.
 A: You seem to be almost there.
Let $\phi$ be an automorphism of $G$. We wish to show that $\phi(K)=K$ (the inclusion suffices in the finite case since $\phi$ is one-to-one, but you never restrict to finite groups, so you need to prove equality).
To your first question: you can't "restrict" $\phi$ to $G/N$, but you can use $\phi$ to define ("induce") a morphism $G/N\to G/N$, as follows.
Lemma. Let $N\triangleleft G$, and let $\psi\colon G\to G$ be a morphism. If $\psi(N)\subseteq N$, then $\phi$ induces a morphism $\overline{\psi}\colon G/N \to G/N$ by $\overline{\psi}(gN) = \psi(g)N$.
Proof. First we show $\overline{\psi}$ is well-defined: if $gN= g'N$, then $g^{-1}g'\in N$, hence $\psi(g^{-1}g')\in N$ (since $\psi(N)\subseteq N$). Therefore, $\psi(g)^{-1}\psi(g')\in N$, so $\psi(g)N = \psi(g')N$. Thus, $\overline{\psi}(gN) = \psi(g)N = \psi(g')N = \overline{\psi}(g'N)$, proving that $\overline{\psi}$ is well-defined.
Second, we verify it is a morphism:
$$\overline{\psi}(xNyN) = \overline{\psi}(xyN) = \psi(xy)N = \psi(x)\psi(y)N = \psi(x)N\psi(y)N=\overline{\psi}(xN)\overline{\psi}(yN).$$
This proves the Lemma. $\Box$
Since $N$ is characteristic in $G$, we know that $\phi(N)=N$. Thus, $\phi$ induces a morphism $\overline{\phi}\colon G/N \to G/N$. So we have that.
Now, in order ot use the fact that $K/N$ is characteristic in $G/N$, we must also make sure that $\overline{\phi}$ is an automorphism of $G/N$. This follows because $\phi$ is an automorphism of $G$: given $gN\in G/N$, we know there exists $x\in G$ such that $\phi(x)=g$, so $\overline{\phi}(xN) = \phi(x)N = gN$, so $\overline{\phi}$ is surjective. And if $gN\in\ker(\overline{\phi})$, then $\phi(g)N = N$, so $\phi(g)\in N$. But since $\phi$ is an automorphism and $\phi(N)=N$ (or because $N$ is characteristic in $G$), we have that this means that $g=\phi^{-1}(\phi(g)) \in \phi^{-1}(N) = N$. Thus, $gN=eN$, so $\overline{\phi}$ is one-to-one. Thus, $\overline{\phi}$ is an automorphism of $G/N$.
That means, because $K/N$ is characteristic in $G/N$, that $\overline{\phi}(K/N) = K/N$.
Therefore, if $k\in K$, then $\overline{\phi}(kN) = k'N$ for smoe $k'\in K$. Therefore, $k'N = \overline{\phi}(kN) = \phi(k)N$, so that  means that $\phi(k)N=k'N$. Therefore, $(k')^{-1}\phi(k)\in N\leq K$, so $\phi(k)\in K$.   This proves that $\phi(K)\leq K$.
As this holds for every automorphism of $G$, it also holds for $\phi^{-1}$, so $\phi^{-1}(K)\leq K$. Therefore, $K=\phi(\phi^{-1}(K))\leq \phi(K)$, giving the converse inclusion and proving equality.
Thus, we conclude that $K$ is characteristic in $G$, as desired.
A: Let $\phi\in G\to G/N$ be canonical.
For any $\alpha\in\mathrm{Aut}(G)$, consider the map $\phi\circ\alpha\in G\to G/N$.  Since $N$ is characteristic, $\alpha$ and $\alpha^{-1}$ preserve $N$.  Thus $N=\ker(\phi\circ\alpha)$ and $\phi\circ\alpha$ factors through $G/N$.  Call the induced map $\tilde{\alpha}\in\mathrm{Aut}(G/N)$.
Since $K/N$ is characteristic, it is preserved by $\tilde{\alpha}$.  In particular, if $k\in K$, then $$(\phi\circ\alpha)(k)=\tilde{\alpha}(kN)\in K/N$$  Thus $\alpha(k)\in\langle K,N\rangle$, where the latter subgroup is generated by all elements of $K$ and $N$.  But since $N\subseteq K$ and the latter is a group, we have $\langle K,N\rangle=K$.
