In $\mathbb{R}^n$, if I have a hyperplane H given by $c^Tx = 1$, then $H$ is $n-1$-dimensional, and so there exists an isometry $f: H\longrightarrow \mathbb{R}^{n-1}$.
Does this mean that I can extend $f$ to be an isometry of all of $\mathbb{R}^n$? In other words, does there exist an isometry $g: \mathbb{R}^n\longrightarrow\mathbb{R}^n$ such that $g$ restricted to $H$ is equal to $f$?
I assume the normal vector of $H$ is such that $||c|| = 1$. Fix an orthonormal basis of $H-c = \{x-c : x\in H \}$ (which is a linear subspace in $\mathbb R^n$), say $e_1,\ldots,e_{n-1}$. Every $x\in H$ can be written as $c + \sum_{i=1}^{n-1} c_i e_i$. Then an isometry $f$ can be written as $$ f(x) = Q(x-c) \quad x \in H $$ where $Q$ is a linear isometry from $H-c$ to $\mathbb R^{n-1}$. Now let $e_n$ be the unit vector that is orthogonal to $span\{e_1,\ldots,e_{n-1}\}$. Then one can extend $Q$ to a $Q'$ defined $span\{e_1,\ldots,e_n\}$ by $Q'(e_k)= Q(e_k)$ for $k<n$ and $Q'(e_n) = e_n$. This way $f$ can be extended to a $g$ as $$g(x) = Q'(x-c). $$ To prove that $g$ is an isometry consider
$$ ||g(x)-g(y)||^2=|| Q'(x-c)-Q'(y-c) ||^2 = || Q'(x-y)||^2 = || Q'(P_{H-c}(x-y)) + P_{e_n}(x-y) ||^2 = || Q(P_{H-c}(x-y))|| + ||P_{e_n}(x-y)||^2 = ||P_{H-c}(x-y)||^2 + || P_{e_n}(x-y)||^2 = ||x-y||^2 $$ since $H-c$ and $e_n$ are orthogonal.