Take the upper sheet $S_+$ of the hyperboloid $$ \Sigma = \{(x, y, z) : x^2 + y^2 - z^2 = -1 \} \, .$$ Then define the matrix $$ \Lambda = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \, , $$ and consider the subgroup $\Gamma \subset \text{GL}(n, \mathbb{R})$ of matrices $A$ for which $$ A^T \Lambda A = \Lambda \, .$$ Pick any of these $A \in \Gamma$, and consider the transformation $\varphi : S_+ \to \Sigma : p \mapsto Ap$. The problem is to show that $\varphi$ indeed maps into $\Sigma$.
Trying to directly figure out how $A \in \Gamma$ looks seems (unnecessarily) complicated, so I imagine the necessary structure can be deduced indirectly from the defining equation $A^T \Lambda A = \Lambda$, and maybe the fact that $\Gamma$ is a group. But I cannot find any nice (algebraic) manipulation to arrive at $\phi(S_+) \subset \Sigma$. I thought if I could prove that $A$ is symmetric, and establish some of its eigenvalues, that would bring me the result eventually, but I'm not sure that's the correct path to proceed. Maybe I'm entirely on the wrong track and missing something obvious!
Could you provide a hint for why $\varphi$ would map into the hyperboloid $\Sigma$?
Note that a point $p$ is on $\Sigma$ is
$$ p^T\Lambda p = -1 \tag{1} $$
All you need to do is to show that if $A\in \Gamma$ then $Ap$ follows the condition in (1):
$$ (Ap)^T\Lambda (A p) = p^T(A^T \Lambda A)p = p^T\Lambda p= -1 \tag{2} $$
So $Ap \in \Sigma$