Background
Let $a,b,c,d$ be nonnegative constants and consider the map $T\colon [0,1]\times[0,1] \rightarrow [0,1]\times[0,1]$ defined by
$$ T(x,y) := \left( \frac{1}{1 + ax + by}, \frac{1}{1 + cx + dy} \right), \quad (x,y) \in [0,1] \times [0,1]. $$
I'm interested in this map (and other similar ones) in the context of monotone dynamical systems. They come up, for instance, as the input to output characteristic of controlled dynamical systems with outputs modeling simple biochemical networks.
Question
By Brower Fixed Point Theorem, $T^2:= T \circ T$ has a fixed point. Now is this fixed point unique? If it is not unique, is there a counterexample? Are there reasonable hypotheses on $a,b,c,d$ which would guarantee that to be true?
Of course I'm also interested in the general finite-dimensional case
$$ T(x_1, \ldots, x_n) := \left( \frac{1}{1 + a_{11}x_1 + \cdots + a_{1n}x_n}, \ldots, \frac{1}{1 + a_{n1}x_1 + \cdots + a_{nn}x_n} \right), $$ as well as replacing the $1$'s in the numerator and denominator by general constants. But I'm hoping to find something conceptual in the simplest case which could then be applied in the general scenario.
Progress
(1) Computer simulations with randomly generated coefficients $a,b,c,d$ seem to indicate that this is true in arbitrary finite dimensions, with any nonnegative $a,b,c,d$.
(2) If $a + c < 1$ and $b + d < 1$, then $T$ is a contraction (with respect to the sum-norm $|(x,y)| := |x| + |y|$. In particular, it has no period-2 points, and so $T^2$ has a unique equilibrium. But these hypotheses seem too restrictive. In fact, taking into consideration where $a,b,c,d$ come from, this is not always true.
(3) This is also true if $b = c$ and $a = d = 0$. This makes me wonder whether there could be some sort of diagonalisation argument. Though I have no idea what to do with the nonlinearity.
(4) I'm also attempting to approach this as a global optimization problem. More precisely, I'm looking at the map $$ (x,y) \longmapsto \|T^2(x,y) - (x,y)\|^2 $$ and tried to compute the Hessian using Maple but haven't gotten anywhere so far.