First of all, let's fix some notation: $$g^n_{\nu}="\text{standard nondegenerate bilinear form on $\mathbb{R}^n$ with index of negativity $\nu$}"$$ $$\mathbb{R}^n_{\nu}="\mathbb{R}^n \text{ equipped with the pseudoriemannian metric $g^n_\nu$".}$$
Let $\mathbb{H}^n$ be the riemannian submanifold of $\mathbb{R}^{n+1}_1$ defined as follows: $$\mathbb{H}^n:\begin{cases}x_0^2-x_1^2-...-x_n^2=1 \\ x_0>0\end{cases}$$ and let $O(n,1)$ be the group of linear isomorphisms of $\mathbb{R}^{n+1}$ that preserve $g^{n+1}_1$. Clearly this group succeeds in preserving the first property that defines $\mathbb{H}^n$ but it fails preserving the second one. Clearly, since elements of $O(n,1)$ are homeomorphisms, they either preserve the two connected components of the hyperboloid or they swap them. So it suffices to impose that an element of $O(n,1)$ maps $(1,0,...,0)$ into $\mathbb{H}^n$ i.e. $$O^+(n,1):=\{A\in O(n,1):a_{00}>0\}.$$ Clearly $O^+(n,1)$ is a subgroup of the isometries of $\mathbb{H}^n$, but I'm having trouble in proving that $O^+(n,1)$ is the whole group of isometries!
For the sphere, the same task was easier because an isometry of the sphere (embedded in standard euclidean space) always extends to an isometry of the euclidea space. Also in the hyperbolic case, this fact should hold (substituting the standard euclidean space with the pseudoeuclidean $\mathbb{R}^n_1$) because (a posteriori!) every isometry is of the form $O^+(n,1)$. But I can't find an easy way to construct this extension.