As for iterating polynomials, there are two general forms of polynomials that we currently know how to iterate. They are polynomials of the form
$$P(x)=a(x-b)^d+b$$
and
$$T(x)=\cos(k\arccos(x))$$
and yes, I know the second one doesn't look like a polynomial, but for integer values of $k$, it is a polynomial where the inverse cosine is defined, and can be extended to other values (accurately, for our purposes) using the hyperbolic cosine function. For more info, see Chebyshev Polynomials.
The iteration formulas for each of these is given by
$$P^n(x)=a^{\frac{b^n-1}{b-1}}(x-b)^{d^n}+b$$
and
$$T^n(x)=\cos(k^n\arccos(x))$$
In case you were wondering how this applies to your problem of finding $f$ given that
$$f(1+n)=(Q\circ f)(n)$$
where $Q$ is a polynomial, here's how: if you assign a value for $f(0)$, say $y_0$, then you can say that
$$f(n)=(Q^n\circ f)(0)=Q^n(y_0)$$
which allows you to find $f:\mathbb Z\to\mathbb Z$.
Unfortunately, your proposed functional equation:
$$ h(y)+h^{-1}(y)=2y+y^2 \tag{1} $$ has no
differentiable solutions near $\,y=0.\,$ Suppose that
$$ h(y) = 0 + a_1 y + O(y^2). \tag{2} $$
Then its inverse function is
$$ h^{-1}(y) = 0 + \frac1{a_1} y + O(y^2). \tag{3} $$
Substituting equations $(2)$ and $(3)$ into $(1)$ gives
$$ h(y)\!+\!h^{-1}(y) = \frac{a_1^2+1}{a_1}y \!+\! O(y^2)
= 2y \!+\! O(y^2). \tag{4}$$
The only solutions for $\,a_1\,$ are $\,a_1=1\,$ and
$\,a_1=-1.\,$ However, we are given that
$$ 0<h(x)<x \tag{5}$$
and thus the only possible value is $\,a_1=1.\,$
Suppose we want more terms in the power series
$$ h(y) = 0 + y + a_2y^2 + O(y^3). \tag{6}$$
The inverse function is now
$$ h^{-1}(y) = 0 + y - a_2y^2 + O(y^3). \tag{7}$$
Adding these two equations together is the equation
$$ h(y) + h^{-1}(y) = 0 + 2y + 0y^2 + O(y^3). \tag{8}$$
This implies that equation $(1)$ can not be satisfied.
There is a possibility that there exists an exponent
$\,e\,$ not an integer such that
$\, h(y) = 0 + y + c y^e + \cdots \,$ which perhaps
should be studied according to comments.
In fact, define $$ g(x) : = \sqrt{h(x^2)}. \tag{9}$$
Then solving for a power series expansion gives
$$ g(x) = x - \frac{x^2}{\sqrt{6}} + \frac{x^3}6 + O(x^4) \tag{10}$$ which implies that
$$ h(x^2) = (x^2+x^4/2)+\frac1{\sqrt{6}}f(x) \tag{11} $$
where
$$ f(x) \!=\! -2 x^3
\!-\!\frac{11x^5}{24} \!+\! \frac{117x^7}{1280}
\!-\!\frac{5491x^9}{110592} \!+\! \frac{156538363x^{11}}{3715891200} \!+\! O(x)^{13}. \tag{12}$$
A Wolfram Language code to calculate $\,g(x)\,$ is
ClearAll[x, g, gx];
gx[3] = x - 1/Sqrt[6]*x^2 + O[x]^3;
Do[g = Normal[gx[n]] + O[x]^(2+n); gx[n+1] = Simplify[
g + (x^2 + (Normal[g]/.x -> g)^2 - g^4 - 2*g^2) * 3 /
((4+n)*x^2*Sqrt[6])],
{n, 3, 6}]
As some comments suggest, it seems that the power series
for $\,f(x)\,$ has zero radius of convergence. This makes
finding properties of it difficult. Perhaps we need to let
$\,x\,$ approach $\,\infty.$ In this case we find that
$\,h(x) \approx \sqrt{x}\,$ with infinite number of other
terms. Using equation $(1)$ we can find the expansion
$$ h(x) = x^{1/2} - 1 + \frac12 x^{-1/4} + \frac1x s(x) \tag{13} $$
where $$ s(x) :=
\sum_{k=2}^\infty 2^{-k}(x^{-\frac32 2^{-k}} -
x^{-2^{-k}} ). \tag{14} $$
This gives moderately good approximations as $\,x\,$ gets large and down to $1.$
One method that leads to equations $(13)$ and $(14)$
is as follows. From equation $(1)$ we immediately get
$$ h^{-1}(y) = y^2 + 2y - h(y) \tag{15} $$
and if $\,y=h(x),\,$ then
$$ x = h(x)^2 + 2h(x) - h(h(x)). \tag{16} $$
We start with an approximation and try to find what
additional term will satisfy equation $(16)$. So we guess
$$ h(x) = x^{1/2} + cx^e + \cdots \tag{17} $$
where $\,\cdots\,$ denotes terms with smaller exponents.
Substitute equation $(17)$ in equation $(16)$ to get
$$ x \!=\! (x \!+\! 2cx^{1/2+e} \!+\! \cdots) \!+\!
2x^{1/2} \!+\! \cdots \!=\! x \!+\! 2x^{1/2}(cx^e \!+\! 1)
+ \cdots. \tag{18} $$
This implies $\,0 = cx^e + 1.\,$ Solving this we get
our next guess as
$$ \,h(x) = x^{1/2} - 1 + cx^e + \cdots. \tag{19} $$
Repeating this process leads to equations $(13)$ and
$(14)$.
The series in $(14)$ appears
to converge but I have no proof, only numerical evidence.
I also have no proof that $(13)$ satisfies the functional
equation $(1)$.
An answer to this question by 'Semiclasical' contains
sequence recursions
$$ u_n = u_{n-1} - j_{n-1} \quad \text{ and } \quad
j_n = j_{n-1} - u_n^2. \tag{20} $$
with the property that $\, u_n = h(u_{n-1})\,$ and
that $\,u_n \to 0^+.\,$ I found that for suitable
starting values of $\,u_1\,$ and $\,j_1\,$ we have
$$ u_n = 6n^{-2} - \frac{15}2n^{-4} + \frac{663}{40}
n^{-6} - \frac{43647}{800}n^{-8} + \cdots. \tag{21} $$
An answer to this question by 'Cesareo' uses recursion to construct a sequence of functions $\,h_n(x)\,$ which seems to converge to a global solution. It would be nice to give a proof of this.
Best Answer
First consider a finit ladder (limited to $n$ rungs) drawing below.
If n is sufficiently large the equivalent resistance $R_0$ is quite undependent from the last resistor i.e. what ever $R_n$ be from $0$ to infinity. This is easy to check with numerical recursive calculus thanks to the formula below.
For an infinite ladder the limit found is : $$\boxed{R_0\simeq 2.485339738238447...}$$
Of course the value found is not analytic but numerical. So my answer isn't a final answer.
Hint : Thanks to ISC (Inverse Symbolic Calculator) one can guess a probable analytic formula : $$R_0=\frac{4}{\ln(5)}\simeq 2.48533973823844724...$$ The councidence is correct for more than 16 digits. The proposed formula can be considered as a good conjecture.
The above result was obtained with my own ISC : https://fr.scribd.com/doc/14161596/Mathematiques-experimentales
Similar studies about ladders made of not only resistors but mixed resistors and capacitors : https://fr.scribd.com/doc/71923015/The-Phasance-Concept