Does a topological space always admit a $\mathbb{Z}_2$-action?

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.

$X=\mathbb R$ easily admits a nontrivial $\mathbb Z_2$-action, but it cannot be free -- we can always find a fixed point with the intermediate value theorem.