Defined $A$ as the set of all the functions $f:\mathbb Z\rightarrow \mathbb R$ exists a function $T:A\rightarrow A$ such that $$ (T^2f)(n)=f(n+1)\text{ for all }n\in\mathbb Z\text{ and }f\in A $$ or not? And if codomain of $f$ is $\mathbb Z$?
Remember that $T^2=T\circ T$ and $T^2f=T\left[T(f)\right]$.
And if $T$ is also linear on $A$ seen as a vectorial space over $\mathbb R$?
Let $h$ be your favorite bijection $\mathbb R^2\to\mathbb R$, and let $\pi_1,\pi_2$ be the corresponding projections such that $h(\pi_1(x),\pi_2(x))=x$. Then take $$ (Tf)(n) = h(\pi_2(f(n)),\pi_1(f(n+1))) $$
The same construction works for every infinite codomain for the $f$s, assuming the axiom of choice. Things get trickier for finite codomains. For a codomain whose size is a perfect square, a variant of the above construction will work; on the other hand for codomains of size $4n+2$ or $4n+3$ the task is impossible. I am not yet sure of the remaining cases.
There can't be linear $T$ with this property as long as the field is $\mathbb R$ or a subfield of $\mathbb R$ such as $\mathbb Q$. Consider the functions $$ f_1(n) = 1 \qquad f_{-1}(n) = (-1)^n $$
Since $T^2(Tf_1)= T^3f_1 = T(T^2f_1)=Tf_1$ it must be that $Tf_1$ is something that maps to itself under $T^2$, which means that it is a constant function, that is $Tf_1 = \lambda f_1$ for some $\lambda$.
Then how about $Tf_{-1}$? It will need to be something with period $2$ under $T^2$, which means that it is a linear combination $af_1 + bf_{-1}$. But the equation for $T^2 f_{-1}$ becomes $T^2 f_{-1} = -f_{-1}$, and the preceding boils down to the matrix equation $$ \begin{pmatrix} \lambda & a \\ 0 & b \end{pmatrix}^2 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $$ But this is impossible because it would require $b^2=-1$.