Let $G$ be any locally compact group and $H$ be a compact group.
We know that a map $F: G \rightarrow G$ is called affine if there exists some $\alpha \in G$ and an automorphism $\Lambda:G\rightarrow G$ such that $F=\alpha\Lambda$.
A continuous action of $H$ on $G$ is a continuous map $\pi: H\times G \rightarrow G$ given by $\pi(h, g) = h.g$ such that $e.g =g$ for all $g\in G$ and $(hk).g= h. (k.g)$ for all $h, k \in H$, $g\in G$.
Hence a continuous action $\pi$ of $H$ on $G$ is called a continuous affine action if for each $h\in H$, the map $g\mapsto (h.g): G \rightarrow G$ is affine.
My Question: Could you please give me an example of such a continuous affine action where atleast one of the maps $g\mapsto (h.g): G \rightarrow G$ is (of course) affine, but not an automorphism?
Thank you in advance, for your help.
Let $G=(\mathbb{R},+)$ and $H=C_2=\langle t\rangle$. Then we may define: $$t.x=1-x=\Lambda(x)+1,$$ for all $x\in \mathbb{R}$, where $\Lambda(x)=-x$ for all $x\in \mathbb{R}$.
Alternatively, you could take $G=H=S^1\subset \mathbb{C}^\times$. For $g\in G$ and $h\in H$ we can define $h.g=hg=h\Lambda(g)$, where $\Lambda$ is the identity.
For a third example, consider $G=SU(2)\times S^1$ and $H=S^1\subset \mathbb{C}^\times$. Let $\Lambda_h\colon G\to G$ denote conjugation by:$$\left(\begin{array}{cc}h&0\\0&\bar{h}\end{array}\right)$$ on the factor $SU(2)$, and the identity on the factor $S^1$.
Then we may define $$h.g=(1,h)\Lambda_h(g).$$