Topological space with fundamental group $\mathbb{Z}/3\mathbb{Z}$.

How about the unit sphere $S^3$? If $A$ is a $2$-by-$2$ rotation matrix for angle $2\pi/n$, then the block matrix $\pmatrix{A&0\\0&A}$ acts on $S^3$ without fixed points.

There is an explicit construction:

Consider the CW complex given by three $0$ cells, three $1$ cells and a two cell pasted in (this is a filled in triangle.)

Now, considering the circle with a base point, and pasting in this $CW$-complex by identifying all three vertices with the base point (in a three fold way.)

The fundamental group of the resulting space is exactly $\mathbb Z_3$ by construction, since the two cell gives the relation $a^3=1$ on the fundamental group.

This is called the presentation complex by the way.

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If you want to construct the Cayley complex (the universal cover of the presentation complex), you will need to take three disks and paste them to the $1$ skeleton of the first construction. $\mathbb Z_3$ will act by cyclic permutation of the $2$-cells, and taking the quotient will give you exactly the presentation complex. Note that this is a simply connected space and that $\mathbb Z_3$ acts freely on the two cells.