Let $G$ be a topological group and $X$ be a topological space.
Let $\alpha$, $\beta:G\times X\to X$ be two group actions. We say that these two actions are homotopic actions if there is a continuous path of actions connecting $\alpha$ to $\beta$. This means that there is a continuous function $\Gamma:[0,1]\times G\times X \to X$ such that each $\Gamma\_t$ is a group action and we have $\Gamma\_0=\alpha, \Gamma\_1=\beta$.


**Question 1:** What is an example of the following situation:
 There are two group actions $\alpha$, $\beta$ by $G=S^1$ on a manifold $M$ which are not homotopic actions but for every $g\in G$, the two homeomorphisms $\alpha\_g$, $\beta\_g$ of $M$ are homotopic maps?


**Question 2:** What is an example of the following situation:
 There are two group actions $\alpha$, $\beta$ by $ G=S^1$ on a manifold $M$ which are not homotopic actions but $\alpha, \beta$ are homotopic maps as maps from $G\times X$ to $X$?



Homotopy of group actions
It's enough to pick a contractible manifold $M$ with two non-homotopic actions. For example, let us pick $M=\mathbb{R}^2$ with two actions of $S^1$, the trivial one and the one given by rotations. These two actions are not homotopic. In fact for every action $\rho$ of $S^1$ on $\mathbb{R}^2$ we can consider the map $\mathbb{R}^2\times S^1\to GL\_2^+(\mathbb{R})$ sending $(x,\lambda)$ to the differential of $\rho(\lambda)$ at $x$. This map is clearly homotopy invariant under the action, and its degree as a map $S^1\to S^1$ is 0 for the trivial action and 1 for the defining action.


However all continuous maps with target $\mathbb{R}^2$ are homotopic.


