Relation between coefficients of expansions Related to Relations between coefficients of expansions of a rational function at 0 and infinity
I commented at the linked question that the question seemed less about what happened "at infinity", and more about what happened "away from zero". And to some extent, the answer confirmed it by discussing the rank of (the biinfinite matrix corresponding to) the biinfinite sequence - which seems to be the degree of the rational function "away from zero and infinity". This is related to the functions $t$ and $t^{-1}$ on $\mathbb{P}^1$; they are the unique functions with a single zero and single pole at $0$ and $\infty$, or vice versa.
So that brings up a corresponding question:
Are there similar algebraic relations that could be made between the expressions of a rational function $f(t)$ when it is expressed as $A(t), B(t) \in F((t))$ such that $f(t) = A(t), f(t - c) = B(t)$ for some constant $c$?
This isn't quite a generalization of the original question, but gets into another interesting situation - when the zeroes don't match, but the poles do. More generally, we can separate the poles as well, leading to:
Are there similar algebraic relations that could be made between $A(t), B(t) \in F((t))$ such that $f(t) = A(t), B(t) = f(\frac{at + b}{ct + d})$?
As in the above question, there is no expectation of a finite algebraic relation. Instead, I'm hoping for a series of "loosening" conditions (similar to the conditions that the biinfinite matrix be rank $n$ for some $n \in \mathbb{Z})$, though I would expect that more than $1$ parameter would be necessary.
 A: Let
\begin{equation}
M = \begin{pmatrix} a & b \\ c & d\\ \end{pmatrix}
\end{equation}
and consider the following transformation
\begin{equation}
\left[B(M )\right]z = \sum_{ k \in \mathbb{Z}} b_k \left[Mz\right]^k
\end{equation}
on $\mathbb{P}^1\mathbb{(F)}$.
What can we deduce from
\begin{equation}
\left[B(M )\right]z = \left[A(I)\right]z =  \sum_{ k \in \mathbb{Z}} a_k \left[Iz\right]^k =  \sum_{ k \in \mathbb{Z}} a_k z^k
\end{equation}
where
\begin{equation}
I = \begin{pmatrix} 1 & 0 \\ 0 & 1\\ \end{pmatrix}?
\end{equation}
If we let $\mathbb{F} = \mathbb{C}  $ then we can assume that $M$ has one eigenvector; furthermore, in terms of projective geometry, the eigenvector(s) are fixed points since
\begin{equation}
Mz = \lambda z = \lambda \begin{pmatrix} x \\ y \end{pmatrix} \implies M \begin{bmatrix} x : y  \end{bmatrix} = \lambda \begin{bmatrix} x : y  \end{bmatrix}  =  \begin{bmatrix} x : y  \end{bmatrix} ,
\end{equation} see [Needham ch .3] for a proof. For example

$a=d = 0 \land b=c=1 \implies \begin{bmatrix} 1 : 1  \end{bmatrix},\begin{bmatrix} -1 : 1  \end{bmatrix} $
  are fixed points

(i.e. $M \equiv \frac{1}{t}$ has $\pm 1$ as fixed points); this is perhaps a rigorous restatement of your comment

"seemed less about what happened 'at infinity,' and more about what
  happened 'away from zero'"

since the defining property of the transformation $M \equiv \frac{1}{t}$ is that it has $\pm 1$ as fixed points as well as the fact that it permutes $\infty$ ($\overset{\text{def}}{=} [1,0]$) with $0$; more generally a Mobius transformation is uniquely defined by its behavior on 3 points (see [Needham ch .3]). 

In particular the Mobius transformation $M \equiv \frac{1}{t}$  is "more about what is happening to" the disc $D_1 = \{|z| < 1 \ | \ z \in \mathbb{P}^1\mathbb{(F)} \}$ and the disc $D_2 = \{|z| > 1 \ | \ z \in \mathbb{P}^1\mathbb{(F)} \}$ on the Rieman sphere $\mathbb{P}^1\mathbb{(F)} $ in the sense that the two discs are being permuted; i.e. $M(D_1) = D_2 $ and $M(D_2) = D_1 $.  

Since a Mobius transformation preserves conics and in particular preserves discs, a "generic" Mobius transformation (case #\ref{3}) satisfies a similar property by that given in the answer to the post you linked. Let us analyze the generating functions and prove it rigorously. 
It's worth considering a degenerate case first 

Case #1 $c = 0$

If $c=0 $ then (after relabeling if necessary) $M (z)= (az+b)$ so that $A(z) = B(M(z))$ implies 
\begin{equation}
\sum_{ k \in \mathbb{Z}} a_k z^k = \sum_{ k \in \mathbb{Z}} b_k (M (z))^k =  \sum_{ k \in \mathbb{Z}} b_k (a z+b)^k = \sum_{ k \in \mathbb{Z}} b_k \sum_{n \in \mathbb{Z}}  \binom{k}{n}(az)^nb^{k-n} 
\end{equation}
\begin{equation}
\sum_{ k \in \mathbb{Z}} a_k z^k= \sum_{ n \in \mathbb{Z}} b_n \sum_{k \in \mathbb{Z}}  \binom{n}{k}(az)^kb^{n-k} = \sum_{ n \in \mathbb{Z}}\left( \sum_{k \leq n } b_n   \binom{n}{k}a^kb^{n-k}\right)z^k
\end{equation}
so that

\begin{equation} \label{1}  \tag{1} A(z) = B(M(z)) \implies \left( \sum_{k \leq n } b_n   \binom{n}{k}a^kb^{n-k}\right)=a_k \end{equation}

The next case 

Case #2 $ad-bc = 0$ 

is trivial since it implies that 

\begin{equation} A(z)= B(M(z)) = B\left(\frac{a}{c}\right) \end{equation}

is effectively a constant function in its domain of convergence (and therefore there is an infinite number of "un-interesting" $B$ that satisfy this equation).

Case #3 $$ad-bc\neq 0 \land c \neq 0\label{3} \tag{3}$$ i.e. $M$ is an "interesting" Mobius transformation.

In this case, we have that that $M$ factors through the following functions


*

*$T_1: z \mapsto z+ \frac{d}{c}$

*$R_1: z \mapsto  \frac{1}{z}$

*$R_2: z \mapsto \frac{ad-bc}{c^2}z$

*$T_2: z \mapsto z+ \frac{a}{c}$
(see equation (3) of [Needham ch .3]) so that 
\begin{equation}
M= T_2 \circ R_2 \circ R_1 \circ T_1
\end{equation}
so that if we let 
\begin{equation}
C(z)=  A\left(T_1^{-1}(z)\right)
\end{equation}
and let 
\begin{equation}
D(z)=  B\left(T_2 \left(R_2 \left(z \right)\right)\right)
\end{equation}
then we have that 

\begin{equation} A(z) = B\left(\frac{az+b}{cz+d}\right) \iff \end{equation}
\begin{equation}  \iff C(z) =
 D\left(\frac{1}{z}\right) \land A(z)= C(T_1 (z)) \land D(z)=  B\left(T_2 \left(R_2 \left(z \right)\right)\right) \end{equation}

as was needed to show. In other words, by applying eq.(\ref{1}) we have that it reduces to the following conditions

  
*
  
*$\left( \sum_{k \leq n } c_n   \binom{n}{k}\left(\frac{d}{c}\right)^{n-k}\right)=a_k $
  
*$\left( \sum_{k \leq n } b_n   \binom{n}{k}\left(\frac{ad-bc}{c^2}\right)^k\left(\frac{d}{c}\right)^{n-k}\right)=d_k$
  
*$C(z)$ and $D(z)$ satisfy the property by that given in the answer to the post you linked

One final remark: the property that was given in the answer to the post you linked is just one of many nice properties that rational generating functions satisfy. The property I proved here has a geometric flavor but there are (as the rigorous proof somewhat suggests) many nice combinatorial properties as well. The whole fourth chapter of Stanley is dedicated to rational generating functions. Right off the bat, he gives theorem 4.1.1 which gives four equivalent conditions on rational generating functions; the contender for most straightforward combinatorial interpretation perhaps being condition (ii) which states 
\begin{equation}
A(x) = \sum_{n\geq 0} f(n)x^n = \frac{p(x)}{q(x)} \iff (\forall n \in \mathbb{N}) \ f(n+d)+\alpha_1f(n+d-1)+...+\alpha_d f(n)  =0
\end{equation}
where $\ \alpha_dx^d+\alpha_{d-1}x^{d-1}+...+\alpha_1 x+1  $ and $\text{deg}(p) \leq \text{deg}(q)$, i.e. that $A(x)$ is the generating function for a linear homogenous recurrence relation.
