TITLE: Is every Lie subgroup of a Lie group isometric to all its conjugates?
QUESTION [7 upvotes]: Let $G$ be a Lie group with a left invariant metric. Assume that $N$ is a Lie subgroup of $G$.
For a given $g\in G$, are $N$ and $g^{-1} N g$ necessarily isometric Riemannian manifold when they inherit the original metric of $G$?
I was inspired by this MSE question:
https://math.stackexchange.com/questions/3121058/almost-normal-subgroup

REPLY [3 votes]: As Francois Ziegler pointed out, it is not true in general. 
The map $\sigma: N \to gNg^{-1}$, $\sigma(n)=gng^{-1}$, is an isometry if and only if 
$$
\langle Ad(g)\cdot X,Ad(g)\cdot Y\rangle=\langle X,Y\rangle
\quad\text{for all $X,Y \in\mathfrak n:=\textrm{Lie}(N)$.}
$$ 
This follows since a left-invariant metric on a Lie group is determined by the inner product on the tangent space at the identity (identified with the Lie algebra), and also because the Lie algebra of $gNg^{-1}$ is $Ad(g)\cdot \mathfrak n$. 
Note that a bi-invariant metric satisfies this condition, but the space of them may be usually larger. 
I am not sure whether there might be an example of an isometry between $N$ and $gNg^{-1}$ when $\sigma$ is not an isometry.