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, 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 $G=\mathrm{GL}(3,\mathbf R)$.

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.