I feel weird when I saw the definition of stabilizing automorphisms of groups. If the diagram is required to be commutative, then isn't the only case of $\varphi$ is that it is nothing but an identity function? Shouldn't the maps $i$ and $p$ in below rows be $i'$ and $p'$(i.e. may be different from the above $i$ and $p$)?
PS: I read the $1_K$ as the identity map on $K$, hopefully I'm not wrong.
Well the identity of $E$ is one stabilizing automorphism, but there may be others.
Let's take as an example $E=\mathbb{Z/4Z}$, $K=2\mathbb{Z/4Z}$, $Q=\mathbb{Z/2Z}$ with the obvious $i$ and $p$; and take $\varphi = -id$.
Then $\varphi \neq id$, but $\varphi_{\mid K} = id_K$ and the induced map on $Q$ is also the identity, therefore $\varphi$ makes the diagram commute but isn't the identity.
If you want, you can define generally morphisms of extensions as triples of maps making the obvious diagrams commute, well here $Stab(Q,K)$ is a subgroup of the group of automorphisms of the extension $K\to E\to Q$