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Edit: According to the comment of Ycor we remove the phrase "Naturally arises from a left invariant metric'

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. We fix an orientation for $G$. We denote by $H$ the group of all orientation preserving diffeomorphisms $f$ of $G$ such that for every $X\in \mathfrak{g}$ we have $f^* X$ belong to $\mathfrak{g}$.

Is there a natural Lie group structure for $H$ which contains $G$ as a closed normal subgroup? How can the constant curvature of $H$ and also its Lie algebra be related to those objects for $G$?

As a simplest example we get the group of affine linear maps $x\mapsto ax+b,\quad a>0$ of the real line when $G=\mathbb{R}$, a model of Poincare upper plane.

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    $\begingroup$ I assume you are including $\mathfrak{g}$ into vector fields on $G$ by taking the left-invariant extension? $\endgroup$
    – user44191
    Commented Sep 9, 2019 at 16:20
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    $\begingroup$ Isn't $H$ just the orientation-preserving subgroup of $Aut_{Lie}(G)$? $\endgroup$
    – user44191
    Commented Sep 9, 2019 at 17:03
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    $\begingroup$ @user44191 is right, and the argument is purely abstract: in a group $G$, denoting by $G_{\mathrm{left}}$ the subgroup of the group $S(G)$ of permutations of $G$ consisting of left translations, then its normalizer in $S(G)$ is the semidirect product $\mathrm{Aut}(G)\ltimes G_{\mathrm{left}}$ (exercise). Restricting to the subgroup of orientation-preserving diffeomorphisms gives the conclusion. $\endgroup$
    – YCor
    Commented Sep 9, 2019 at 18:10
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    $\begingroup$ @AliTaghavi It isn't. Every manifold admits a Riemannian metric, including non-orientable ones... also the isometry group of many Riemannian manifolds, including positive-dimensional Euclidean spaces, does not preserve the orientation. Your argument fails because you're just working in a chart and using the orientation of the chart. $\endgroup$
    – YCor
    Commented Sep 9, 2019 at 19:55
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    $\begingroup$ @YCor I think my answer is incorrect (as shown by the example in the OP), though I was aiming at the "purely abstract" method you have. I should have said that the subgroup of $H$ that fixes the identity is the orientation-preserving subgroup $Aut^+(G)$, which leads to an answer of $Aut^+(G) \ltimes G_{left}$, if I'm not mistaken again. $\endgroup$
    – user44191
    Commented Sep 10, 2019 at 21:35

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