There are numerous ways to compute approximations to $\pi$.
Is it possible to find a mapping $f:A \to A$ where $A \subseteq \mathbb{Q}$ such that the iterates $f^{n}(x)$ tend towards $\pi$ for any starting point $x$?
For example maybe this $f$ is some contractive algorithm such that each application of $f$ improves the approximation.
More generally, is it the case that for any transcendental number there is a map of a subset of the rationals to itself whose iterates $f^{n}(x)$ converges to that number?
Yes.
For each rational number $a$, let $f(a)$ be any rational number $b$ such that $\vert b-\pi\vert< \frac{\vert a-\pi\vert}{2}$. And of course this works for any irrational in place of $\pi$. (The "$1\over 2$" is needed to make sure that $f^n(x)\rightarrow \pi$, rather than just $\vert f^{n+1}(x)\vert<\vert f^n(x)\vert$.)
We can even get rid of the apparent arbitrariness in the definition of $f$, above. The rationals are well-orderable - there is a linear order $\prec$ on $\mathbb{Q}$ (not the usual ordering) such that every nonempty subset of $\mathbb{Q}$ has a $\prec$-least element. (In fact, $\mathbb{Q}$ is countable, but we don't need that.) This lets us define such an $f$ explicitly as $$f(a)=\mbox{the $\prec$-least element of $\{b\in\mathbb{Q}: \vert b-\pi\vert<{1\over 2}\vert a-\pi\vert\}$}.$$ (It's a good exercise to come up with such a $\prec$.)