Determine whether $\phi$ is an automorphism Let $H$ be a proper subgroup of $G$ and let $\psi$ be an automorphism of $H$ other than the identity mapping. Define a mapping $\phi:G\rightarrow G$ by  
$\phi(x)=$\begin{cases}\psi(x) \text{ if } x\in H\\x \text{ if } x\not\in H\end{cases}
Is $\phi(x)$ an automorphism?

Take $g \in G\backslash H$. Every element in $H$ is a product of two elements  from $G \backslash H$. 
$$h = g^{-1} \cdot gh$$. 
If the morphism $\phi$ conincides with $\mathbb{1}_G$ on $G \backslash H$ then it is $\mathbb{1}_G$
