Recently I have been reading a lot about $\mathbb{Z}_2$-actions on topological spaces. Mainly I was focused on surfaces such as the sphere, torus and Klein bottle and here the existence of a nontrivial $\mathbb{Z}_2$-action is rather simple. But I was wondering if a general topological space always admits a nontrivial continuous $\mathbb{Z}_2$-action? If not, then more specific, does a manifold always admit a nontrivial continuous $\mathbb{Z}_2$-action?
For a manifold $M$ I was thinking about the fact that we can embed $M$ into $\mathbb{R}^N$ for some $N >0$ and then $M$ can inherit a $\mathbb{Z}_2$-action from $\mathbb{R}^N$ but then when one looks at the spiral in $\mathbb{R}^2$ we see that this spiral does not inherit for example the antipodality of $\mathbb{R}^2$.
Extra: I was also wondering that if there are spaces that do admit a nontrivial continuous $\mathbb{Z}_2$-action, do these space then also admit a free $\mathbb{Z}_2$-action? By free I mean that the action is fixed point free.
If anyone knows some basic examples that do not admit a continuous (free) $\mathbb{Z}_2$-action. Please do share. I seem to be unable to find one.
Thank you in advance!
Consider $X=[0,1)$. By connectedness arguments it follows that $0$ is fixed by every homeomorphism and as a consequence every $\mathbb{Z}_2$-action on $X$ is trivial (to verify this, show that every subset of the form $[0,\varepsilon]$ gets fixed).
If we consider manifolds then I can't give you examples that do not admit a non-trivial $\mathbb{Z}_2$-action (which do not consist of only one point). However, if we consider free actions then the quotient is a manifold and we obtain a covering. It is a consequence of the Lefschetz fixed point theorem that the only free actions on $S^{2n}$ are trivial or given by $\mathbb{Z}_2$. But if $\mathbb{R}P^{2n}$ had a free $\mathbb{Z}_2$-action, then the quotient would have a fundamental group of order $4$ with universal covering $S^{2n}$, which is not possible.