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I need an example of a topological space $ X$ such that $S_3$ acts freely on $X $. I tried to find an easy example with google, but I didn't suceed.

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    (Why the 'algebraic-geometry' tag?) Any example will do? Let $S_n$ permute the coordinates of $\{ (x_1,\ldots,x_i,\ldots,x_j,\ldots,x_n) \mid x_i \ne x_j \text{ for }i\ne j \}$. – 2017-01-10

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$S_3$ can be identified with the group of unit quaternions. That gives $S_3$ the structure of a topological group and allows a free action on itself by group multiplication. By the same token there is a free action $S_3 \times \mathbb R^4 \to \mathbb R^4$ obtained by treating $\mathbb R^4$ as the space of all quaternions.

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    You need to throw away (0,0,0,0) first. – 2017-01-10
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    Oh yeah, you have to consider the space being acted upon as $\mathbb R_{\ne 0}$ to ensure it is free. This problem doesn't arise if $S_3$ itself is the space being acted upon. – 2017-01-10
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    $S_3$ is not the sphere. – 2017-02-11
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Consider $S_3$ as a topological group under the discrete topology and take the action by left multiplication on itself.

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    ... which also works for every topological group. – 2017-01-10
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Let

$X=SL_6(\mathbb{R})$ a simply connected topological space

and

$G=${$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1\\\end{bmatrix}$, $\begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1\\\end{bmatrix}$, $\begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0\\\end{bmatrix}$,

$\begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0\\\end{bmatrix}$, $\begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0\\\end{bmatrix}$,$\begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0\\\end{bmatrix}$}.

It's easy to see that every element of the group $G$ is a block matrix. On the main diagonal is the identity matrix with lines changed under the action of a permutation from $S_3$.

Therefore , $G\simeq S_3$. The action of $G$ on $X$ is without fixed points. In this way I obtain that $\pi_1(X/G)= S_3$.