What is the solution for these two equalities $ x = \lambda(Ax) $ and $ x^TAx=1 $? Let $A \in \mathbb{R}^{4 \times 4}$ be a symmetric matrix with two distinct positive eigenvalues and other eigenvalues of $A$ are nonpositive.
What is the solution for $x$ when $\lambda >0$
$$ x = \lambda(Ax) $$ and $$ x^TAx=1  $$

Suppose that $x=\lambda Ax$, then
$$\lambda Ax-x=0$$
$$\lambda(Ax-\lambda^{-1}x)=0$$
$$(A-\lambda^{-1}I)x=0$$
So $x$ is an eigenvector with eigenvalue $\lambda^{-1}>0$. Moreover,
$$x^T Ax=\lambda^{-1}x^T x=1 \Rightarrow x^T x=||x||^2=\lambda$$
So any solutions, $x$, must be eigenvectors with (positive) eigenvalue $\lambda^{-1}$ and norm $||x||=\sqrt{\lambda}$. If the unit eigenvectors of $A$ are $v_1$ and $v_2$ 
 with the corresponding positve eigenvalues $\mu_1$ and $\mu_2$, then the solutions to your system of equations are:
$$(x,\lambda) \in \{(\sqrt{\mu_1^{-1}}v_1,\mu_1^{-1}),(\sqrt{\mu_2^{-1}}v_1,\mu_2^{-1}) \}$$
