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I want to give a talk on group action on topological spaces.So i was searching for some interesting group action on spaces.Maybe with some geometric view or related to algebraic topology or in general.It will be very helpful if someone suggest some good examples.

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    No, it doesn't rely on C*-algebras, @BerrickCalebFillmore. Given $\alpha$ define $h(x) := \alpha(1, x)$, which is continuous because $\alpha$ is. This $h$ is a homeomorphism because its inverse is the continuous function $g(x):=\alpha(-1,x)$ ---to check it is the inverse you just use that $\alpha$ is an action. Checking that $\alpha(n,x) = h^n(x)$ again just uses that $\alpha$ is an action. (You also don't need $X$ to be locally compact or Hausdorff.) – 2017-02-28
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    @Omar: Hi Omar. Thanks for pointing that out. What I meant to say is, given a locally compact Hausdorff topological space $ X $, if $ \alpha $ is an action of the group $ \mathbb{Z} $ on the $ C^{\ast} $-algebra $ {C_{0}}(X) $ by $ \ast $-automorphisms, then there exists a self-homeomorphism $ h: X \to X $ such that $ \alpha(n,f) = f \circ h^{n} $ for any $ f \in {C_{0}}(X) $ and $ n \in \mathbb{Z} $. – 2017-04-21

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When testing embeddability of a simplicial complex $X$ into $\Bbb R^n$, one obvious necessary condition is the existence of a $\Bbb Z_2$ equivariant map $(X\times X)/\Delta\,\to S^{n-1}$ where the $\Bbb Z_2$ action on the first space is $(x,y)\mapsto (y,x)$. This is because any embedding $f$ gives rise to the map $$(x,y)\mapsto \frac{f(x)-f(y)}{\|f(x)-f(y)\|}.$$

In some range of dimensions, this condition is even sufficient (see Haefliger-Weber theorem).

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I like to give the example of the action of $C_2$, the cyclic group of order $2$, on the circle $S^1$, the set of complex numbers $z$ of modulus $1$, by $z \mapsto \bar{z}$. The quotient of $S^1$ by the action is homeomorphic to the semicircle, which is contractible. However the action on a fundamental group, say $\pi_1(S^1,1)\cong \mathbb Z$, is $n \mapsto -n$, and the quotient by that action is $\mathbb Z_2$. What has gone wrong?

There are of course two fixed points of the action, namely $\pm 1$. So we need to take the fundamental groupoid $\pi_1(S^1, \{1,-1\})$ on the set $\{\pm 1\}$ of base points. The quotient of this groupoid by the induced action is correct!

The general situation (i.e. considering orbit groupoids) is covered in Topology and Groupoids, Chapter $11$, and, it seems, in no other topology text.