Last year, in a talk of Michel Waldschmidt's, I remember hearing a statement along the lines of the title of this question, that is, "The Galois group of $\pi$ is $\mathbb{Z}$.". In what sense/framework is this true? What was meant exactly  and can this notion be made precise?

6$\begingroup$ Crossposted from math.SE: math.stackexchange.com/q/94994 $\endgroup$ – Theo Buehler Dec 30 '11 at 2:52

7$\begingroup$ See this nice survey by Andre: arxiv.org/abs/0805.2569 $\endgroup$ – user18237 Dec 30 '11 at 3:04

9$\begingroup$ @JSeaton: When you crosspost to/from math.SE, please inform readers (in both sites) of the fact; this helps prevent duplication of efforts. If you get a good answer in one site, please be sure to note it in the other. $\endgroup$ – Arturo Magidin Dec 30 '11 at 4:29

$\begingroup$ @Arturo: Okay, noted. $\endgroup$ – Joshua Seaton Dec 30 '11 at 16:31
I gather that the idea behind $\mathrm{Gal}(\pi)=\mathbb{Z}\backslash\{0\}$ (not $\mathbb{Z}$, $0$ is not a conjugate of $\pi$!) comes from Euler's formula:
$$\prod_{n\in \mathbb{Z\backslash\{0\}}}\bigg(1\frac{x}{n\pi}\bigg)=\frac{\mathrm{sin}(x)}{x} \in\mathbb{Q}\{x\} $$
which can make you think of those $n\pi$ as the conjugates of $\pi$.
But in order for the Galois groups to act transitively on the conjugates you need all the nonzero rationals, so that $\mathrm{Gal}(\pi)=\mathbb{Q^{\times}}$.
For some actual evidence that $\mathrm{Gal}(\pi)=\mathbb{Q^{\times}}$ is the right answer (this is, consistent with the conjectural picture of periods and motives), see sections 3 and 5 of Galois theory, motives and transcendental numbers by Yves Andre, alredy mentioned in the comments.

2$\begingroup$ You mean $\mathbb{Z}\setminus\{0\}$ and not $\mathbb{Z}^\times=\{\pm 1\}$. $\endgroup$ – Johannes Hahn Feb 14 '15 at 15:37

$\begingroup$ @JohannesHahn. Yes, by $K^{\times}$ I mean the nonzero elements of $K$. $\endgroup$ – Myshkin Feb 14 '15 at 15:47

14$\begingroup$ Which is nonstandard notation. $k^\times$ is standard notation for the unit group of $k$. If $k$ is a field, then of course this coincides with $k\setminus\{0\}$. But $\mathbb{Z}$ isn't a field. $\endgroup$ – Johannes Hahn Feb 14 '15 at 16:57