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In Euclidean geometry $\mathbb{E}^2(\cong \mathbb{C})$, the group $G=\{z\mapsto raz+b, z\mapsto ra\bar{z}+b\colon a,b\in \mathbb{C},|a|=1, r\in \mathbb{R}^{+} \}$ is precisely the group of similarities of $\mathbb{E}^2$. This group is also a set of bijective maps on $\mathbb{E}^2$, which maps parallel (intersecting) lines to parallel (intersecting) lines.

Question: If we consider $\mathbb{E}^2$ from axiomatic view point, then, is this group precisely the group of bijective morphisms of the geometry?

(My question may not be clear to someone; but what I want to ask is the following: there are points and lines in $\mathbb{E}^2$; and two distinct lines can have either zero or one common point. If we consider the group of bijective maps on $\mathbb{E}^2$ which preserves this coincidence, then is this group precisely the group $G$ mentioned above? If one modifies the question precisely, I will be happy.)

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You're asking if these are all automorphisms of $\mathbf E^2$ as a model for the theory of euclidean geometry? Note that in general, a bijective morphism is not an automorphism. – tomasz Aug 12 '12 at 12:14
An automorphism should preserve inequalities between angles and lengths as well as incidence, I think... But that may follow from incidence-preserving. – tomasz Aug 12 '12 at 12:28

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