Suppose $\rho\in(0,1)$ and $\tau>0$, and consider the sequence of functions $\{g_0(z),\ldots,g_{n+1}(z)\}$ such that, \begin{align} g_0(z)&= \tau(1-\rho z), \\[1.5ex] g_{k }\left( z\right)& =\frac{1+g_{k-1}\left( \lambda_{k}\right) g_{k-1}\left( z\right) }{g_{k-1}\left( \lambda_{k}\right) -g_{k-1}\left( z\right) }\frac{z-\lambda_{k}}{1-\lambda_{k} z}. \end{align} Suppose that $n\ge 2$, $\lambda_{n+1}=0$, and that $\{\lambda_1,\ldots,\lambda_{n}\}$ satisfy \begin{equation}\tag1 \frac{g_{n+1} (0) g_{n+1}(\lambda_j^{-1}) + 1}{g_{n+1}(0) g_{n+1}(\lambda_i^{-1}) +1}\;\; \frac{g_{0}(\lambda_j) {g }_{0}(\lambda_j^{-1})+1}{g_{0}(\lambda_i) {g }_{0}(\lambda_i^{-1})+1} =\frac{g_0(\lambda_j)}{g_0(\lambda_i)}, \;\;\;\text{for } i,j\in \{1,\ldots, n\}. \end{equation}
Conjecture: $|\lambda_i|>1$ for at least one $i\in\{1,\ldots, n\}$.
Observations:
Let $x\equiv\rho +\frac{1+\tau^{-2}}{\rho}$ and notice that $x>2$. We have been able to establish the following partial results:
For $n=2$, equation $(1)$ implies $$x^2-(\lambda_1+\lambda_2)x+\lambda_1\lambda_2=1.$$ It follows that $(x-\lambda_1)(x-\lambda_2)=1$, so that it must be that either $|\lambda_1|>1$ or $|\lambda_2|>1$;
For $n=3$ and $n=4$, equation $(1)$ implies the quadratic equations in this question, and the final step of the result is established in the answers to the question;
For $n= 5$ we have checked the result numerically. We can provide the associated polynomial if someone is interested in it.
One potentially useful property that follows from the definitions is that $$g_{k}( z) -g_{k}(\lambda_{k+1})=a_k \frac{z-\lambda_{k+1}}{1-\lambda_{k}z} $$ for some scalar $a_k$ that depends on $\{\rho,\lambda_1,\ldots,\lambda_{n}\}$.
We have been struggling with this problem for a while, but aside from the particular cases listed above, which we obtained by brute force, we haven't been able to make much progress. Any help would be much appreciated.