The second fundamental form and isometry. What is the effect on the second fundametal form of asurface of applying an isometry of $\Bbb R^3$ ? Or a dilation? 
I posted its answer. 

This answer is not understandable for me in general. Please is someone explains it, I'll be happy. Thank you. 
 A: Let $r=r(u,v)$ be a parametrization of a given surface and let $n=n(u,v)$ denote the normal vector. We use the notation
$$II:=Ldu^2 + 2Mdudv + Ndv^2$$
where $L = r_{uu}\cdot  n$; $M = r_{uv}\cdot  n$; $N = r_{vv}\cdot  n$ for the second fundamental form.
In this answer I proved that the first fundamental form is invariant under 
$r\mapsto r':= Or+b$, where $O$ is a $3\times 3$ orthogonal matrix and $b$ real. Note that $r'=r'(u,v)$. and
$$r'_j=\sum_{k=1}^3 O_{jk}r_k+b_j, $$
for all $j=1,2,3$.
Then $r'_u:=\frac{\partial r'}{\partial u}=O\frac{\partial r}{\partial u}$ and $r'_v:=\frac{\partial r'}{\partial v}=O\frac{\partial r}{\partial u}$, as $O$ does not depend on $(u,v)$; similarly  $r'_{uu}=r_{uu}$ and $r'_{vv}=r_{vv}$. The normal vector w.r.t. $r'$ is then
$$n'=r'_u \times r'_v=\pm n,$$
as proved in the original reference (prop. A.1.6). It follows that
$$L'=  r'_{uu}\cdot  n'=\pm r_{uu}\cdot n=\pm L,$$
$$M' = r'_{uv}\cdot  n'= \pm r_{uv}\cdot n=\pm M,$$
and similarly for the remaining coefficient of the second fundamental form, i.e. $N$.
The case of dilations is treated analogously. Please check this answer again for the explicit computations.
A: For the second part of the question, to understand the effect on the second fundamental form of  applying a dilation, lets consider a hypersurface in $\mathbb{R}^n$ with the metric induced by the usual inner product
$\langle \cdot , \cdot \rangle$.
Then we have a parameterisation $F$ of a surface in $\mathbb{R}^n$, and the dilation
$\tilde{F} = \psi F$, where $\psi$ is a constant (independent of the spacial coordinates).
We then notice that
\begin{equation}
\frac{\partial \tilde{F}}{\partial x_i} = \psi\frac{\partial F}{\partial x_i}, \quad \mbox{and}\quad 
\frac{\partial^2 \tilde{F}}{\partial x_j\partial x_i} = \psi\frac{\partial^2 F}{\partial x_j\partial x_i},
\end{equation}
from which follows
\begin{equation}
\tilde{g}_{ij} = 
\left\langle \frac{\partial \tilde{F}}{\partial x_i} , \frac{\partial \tilde{F}}{\partial x_j} \right\rangle =
\left\langle \psi \frac{\partial F}{\partial x_i} ,\psi \frac{\partial F}{\partial x_j} \right\rangle = 
\psi^2 g_{ij}.
\end{equation}
Since the unit normal vector $\nu$ of the hypersurface $F$ is also a normal unit vector for
the hypersurface $\tilde{F}$, we compute the second fundamental form as
\begin{equation}
\tilde{h}_{ij} = \left\langle \frac{\partial^2 \tilde{F}}{\partial x_j\partial x_i} , \nu\right\rangle = 
\psi \left\langle \frac{\partial^2 F}{\partial x_j\partial x_i} , \nu\right\rangle  =
\psi h_{ij}.
\end{equation}
