Construct non trivial group endomorphism (rigid motion group) My question, in it's general formulation, is : is there a way to construct non trivial group endomorphism other than conjugation ?
Now for my specific needs, I wont to find some endomorphism other than conjugation of the following group :
$$G=\left\{ \left(\begin {array} {c c} a & b \\ 0 & 1
\end {array} \right ) \equiv [a,b] \,:\, a\in U(1) , ~ b\in \mathbb C \right\} \simeq U(1)\ltimes \mathbb C $$
This group is isomorphic to the special Euclidean group $ SE(2)= SO(2)\ltimes \mathbb C $, also called the rigid motions group, or displacement group which is the subgroup of Euclidean group that contains all orientation-preserving isometries (translations and rotations).
For information, the conjugation gives the following endomorphism
$ \mathfrak{h}=[\alpha_{\mathfrak{h}},\beta_{\mathfrak{h}}] $ be a fixed element in $ G $, 
$$\begin{equation}
 \rho_{\mathfrak{h}}(g):= \mathfrak{h} g\mathfrak{h}^{-1}=[a,(1-a)\beta_{\mathfrak{h}}+b\alpha_{\mathfrak{h}}]; \qquad g=[a,b]\in G
\end{equation}$$
 A: In full generality, no.  Any group homomorphism $\phi\colon G\to H$ yields a normal subgroup of $G$ via its kernel (and identifies the morphism up to an isomorphism of the image).  This goes for group endomorphisms, in particular.  So in a simple group the only endomorphisms are the trivial morphism and the automorphisms.  Discarding the trivial morphism, we are left with automorphisms.  If there is an outer automorphism, you are done.  A simple group need not, in general, possess an outer automorphism, however.
So to solve your particular case, start considering the outer automorphisms, and then failing that the normal subgroups and whether $G$ contains a copy of the quotient by said subgroup.  For normal subgroups, one should be staring you in the face from the groups you've listed it as isomorphic to.
Indeed, $SE(2)$ is normal (of index 2) in the full Euclidean group, so conjugation by a reflection acts as an order 2 isomorphism of $SE(2)$.  Do you know if that is an inner isomorphism?
A: For the specific case of the given group $G$, there seem to be at least a few interesting choices:


*

*As usual we can view $U(1)$ as the unit circle in $\Bbb C$, which in particular is preserved (as a set) under complex conjugation. In the bracket notation we can write the multiplication operation as
$$[a, b] \cdot [c, d] := [ac, ad + b];$$
since complex conjugation is a field automorphism, we see immediately that the map $C: G \to G$ defined by $$C: [a, b] \mapsto [\bar{a}, \bar{b}]$$ is a group isomorphism.

*As indicated by the semidirect product notation, the group $N := \{1\} \times \Bbb C < G$ is normal, and the quotient map $\pi: G \to G / N \cong U(1)$ is a homomorphism. Postcomposing with the inclusion homomorphism $\iota$ back into the semidirect product gives the map $$\Pi := \iota \circ \pi$$ defined by $$\Pi : [a, b] \mapsto [a, 0],$$ which neither trivial nor an isomorphism.

*One can combine the above examples to produce the homomorphism $$\Pi \circ C = C \circ \Pi : [a, b] \mapsto [\bar{a}, 0],$$ which again is neither trivial nor an isomorphism.

*One can modify the examples in (2) and (3) by inserting in the composition the power map homomorphism $p_n : U(1) \to U(1)$ defined by $p_n(z) := z^n$ for any $n \in \Bbb Z$ to produce the maps $$P_n := \iota \circ p_n \circ \pi : [a, b] \mapsto [a^n, 0].$$
The map $p_0$ is trivial (and hence so is $P_0$), $p_1$ is the identity map (so $P_1 = \Pi$), and $p_{-1}$ is the conjugation map (so $P_{-1} = C \circ \Pi$), and hence this example subsumes (2) and (3) above. Note also that $P_{-k} = P_k \circ C = C \circ P_k$.
