If I have an action $\theta$ on a set $X$, $\theta : G \times X \to X$ , and I have a map $f: X \to Y$, what conditions do I need to ask to $G$, $X $ and $f$ to guarantee that there exists an action $\phi: G \times Y \to Y$ such that $f$ is equivariant?
In particular, I'm working with $X, Y$ topological spaces.
In the case when $f$ is a homeomorphism the answer is just the canonical action $$\phi(g,y) = \theta(g,f^{-1}(y)),$$ but for more general cases I don't know when the action $\phi$ exists.
In general, it's sufficient -- more than sufficient! --- for $f$ to be bijective; in that case, your canonical action makes $f$ equivariant. Of course, if $f$ is merely a bijection, and not continuous, then the action may well not be continuous, which may be a problem.
You want injectivity because otherwise $\theta(g, f^{-1}(y))$ may not make sense -- it might mean $k$ different things, if $f^{-1}(y)$ has cardinality $k$. OK, it could be weaker -- you could insist that for every $y$, $f^{-1}(y)$ is a subset of some orbit of $\theta$ -- but that seems likely to be hard to check.
You probably WANT surjectivity so that $\phi$ is actually defined for all $y$; if $f^{-1}(y)$ is empty, you don't know where to send $y$. You can, of course, define it any way you like, and things will still be equivariant, but...getting $\phi$ continuous in that case could be really tough.
So I think bijection is a pretty natural sweet-spot. Someone else may be able to correct me, though.