Quotient of a variety and orbits Suppose a group G acts on a variety X and a quotient exists, that is, we have a variety Y and a regular map $\pi : X \rightarrow Y$ so that any regular map $\varphi :X \rightarrow Z$ to another variety Z factors through $\pi$ if and only if $\varphi (p) = \varphi (g(p)) \forall p \in X, g \in G$.
I'm trying to prove that the points of Y correspond to orbits of G on X, i.e.
$\pi (p) = \pi (q) \iff \exists g \in G: g(p) = g(q) $
However, I am stuck. The only triviality I was able to show is that, assuming $\pi (p) = \pi (q)$, we'd have $\pi (g_1(p)) = \pi (g_2(q)) \forall g_1, g_2 \in G$. I guess it boils down to choosing the right variety Z and then make use of the fact that Y is a quotient, but I don't know how. I'd be grateful for any hints.
EDIT: I just realized I have a bad typo in this post. $g(p) = g(q)$ should be $g(p) = q$ , sorry!!
 A: Here's another perspective:  suppose $G$ acts on $X$ and that a variety $Y$ exists which parameterizes the orbits of $G$ on $X$, with a regular morphism $\pi: X \rightarrow Y$ such that $\pi(x') = \pi(x) \Leftrightarrow x' = g \cdot x$ for some $g \in G$.  Then, since any point $y \in Y$ is closed in the Zariski topology and $\pi$ is a continuous map, $\pi^{-1}(y)$ -- an orbit of $G$ in $X$ -- must be closed.
So if $G$ acts on $X$ with non-closed orbits, it cannot have a quotient in the category of algebraic varieties whose points corresond to orbits on $X$.  In QiL's example, the orbit $k^*$ is not closed in $\mathbb{A}^1_k$.
A: By construction, $g(p)=q$ implies that $\pi(p)=\pi(q)$. But the converse is false in general. For example, suppose $k$ is an algebraically closed field, and let $G=k^*$ act on $X=\mathbb A^1_k$ by $(\lambda, z)\mapsto \lambda z$. Then the quotient $X\to Y$ is just the structural morphism $X\to Y=\mathrm{Spec}(k)$ (see below). So $\pi$ is constant. But $0, 1\in \mathbb A^1_k$ are not in the same orbit.
Computation of $X/G$. Let $f: X\to Z$ be any $G$ invariant morphism. Then $f(k^*)$ is one point because $G$ acts transitively on $k^*\subset X$. By continuity of $f$, $f$ is constant. So $f$ factors through the structural morphism $\rho$ of $X$. Therefore $X\to Y$ is equal to $\rho$.  
If $G$ is a finite group acting on a quasi-projective variety $X$, then $G$ act transitively in the fibers of $X\to Y$. See Mumford, Abelian varieties, p 55. 
A: Sadly I still can't comment because I lost the other account.
So the counterexample holds? It does convince me, however I am confused because I got the idea that the converse should be true from p. 123 in Harris' "Algebraic Geometry, A First Course".
He mentions that this "factorization property" I described in the beginning of my question is stronger than:
There exists $\pi: X \rightarrow Y $ surjective so that: $\pi (p) = \pi (q) \iff \exists g \in G: g(p) = q$
So in case I didn't misunderstand the text, the converse should be true. Your construction is presented one page later, but as a counterexample to the statement that  a quotient always exists: This supposedly does not hold here because "[...] there does not exist a surjective morphism from $\mathbb{A}^1$ onto a variety with two points".
I am an absolute beginner in the field of AG. Are there maybe some definitions of quotients which are not equivalent or did I just get something terribly wrong?
