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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 $\mathbb{R}^2\times S^1\to S^1$ sending $(x,\lambda)$ to the determinant of 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.

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.

It's enough to pick a contractible manifold $M$ with two non-homotopic actions. For example, let us pick $M=\mathbb{R}^n$ with two actions of $C_2$, the trivial one and one given by a reflection. These two actions are not homotopic, since in the nontrivial action, the nontrivial element of $C_2$ is sent to a nonzero component of $\mathrm{Diff}(M)$ (if you want, it is not orientation-preserving), but of course all continuous maps with target $M$ are homotopic.

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 $\mathbb{R}^2\times S^1\to S^1$ sending $(x,\lambda)$ to the determinant of 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.

It's enough to pick a contractible manifold $M$ with two non-homotopic actions. For example, let us pick $M=\mathbb{R}^n$ with two actions of $C_2$, the trivial one and one given by a reflection. These two actions are not homotopic, since in the nontrivial action, the nontrivial element of $C_2$ is sent to a nonzero component of $\mathrm{Diff}(M)$ (if you want, it is not orientation-preserving), but of course all continuous maps with target $M$ are homotopic

It's enough to pick a contractible manifold $M$ with two non-homotopic actions. For example, let us pick $M=\mathbb{R}^n$ with two actions of $C_2$, the trivial one and one given by a reflection. These two actions are not homotopic, since in the nontrivial action, the nontrivial element of $C_2$ is sent to a nonzero component of $\mathrm{Diff}(M)$ (if you want, it is not orientation-preserving), but of course all continuous maps with target $M$ are homotopic.