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>



Is every Lie subgroup of a Lie group isometric to all its conjugates?
Unless I miscomputed, the left-invariant metric
$Q(dg,dg)=\operatorname{Tr}\bigl(\overline{g^{-1}dg}\,g^{-1}dg\bigr)$ (bar $=$ transpose) on
\begin{equation}
G=\left\{g=\begin{pmatrix}a&b&c\\0&1&e\\0&0&1\end{pmatrix}:
\begin{matrix}a>0,\\b,c,e\in\mathbf R\end{matrix}\right\},
\qquad
N=\left\{n=\begin{pmatrix}a&b&0\\0&1&0\\0&0&1\end{pmatrix}:
\begin{matrix}a>0,\\b\in\mathbf R\end{matrix}\right\}
\end{equation}
provides a counterexample. Indeed, taking $g=\smash[b]{\begin{pmatrix}1&0&c\\0&1&e\\0&0&1\end{pmatrix}}$ and following Milnor ([1976](//ams.org/mathscinet-getitem?mr=425012), pp. 303, 312–314), one finds that the metric
\begin{equation}
\ \\
(\operatorname{Ad}\_g^\*Q)(dn,dn)=(a^{-1}da\quad a^{-1}db)
\begin{pmatrix}1+c^2&ce\\ce&1+e^2\end{pmatrix}
\begin{pmatrix}a^{-1}da\\a^{-1}db\end{pmatrix}
\end{equation}
(restricted to $N$) has scalar curvature $\ -\dfrac{1+e^2}{1+c^2+e^2},\ $ which depends on $g$.




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**Added:** For simpler, one could of course let $e=0$ throughout, or do this inside $G=\mathrm{GL}(3,\mathbf R)$.


