Totally geodesic subgroups in Lie groups Let $G$ be a Lie group with a left invariant metric $g$.
Let $H$ be a (closed) Lie subgroup of $G$, and assume $g$ is right-$H$-invariant. (That is $d(R_h)_e:T_eG \to T_hG$ is an isometry for every $h \in H$). Note that this is equivalent to the statement, that for every $h \in H$ the map of right multiplication by $h$, viewed as a map $G \mapsto G$ is an isometry.
(This appears to be stronger thant requiring it will be an isometry as a map $H \mapsto H$).
Is it true that $H$ must be totally geodesic in $G$?
Since $g$ is bi-$H$-invariant, the geodesics of $H$ are the one-parameter subgroups of $H$. Hence, the question amounts to:
Are the one-parameter subgroups of $H$ geodesics in $G$?
 A: Edit. I totally misunderstood the question, maybe now its better.
Assume that $g$ is left-$G$ and right-$H$-invariant. Then one can construct a Riemannian submersion $p\colon G\to G/H$. This is now a family of Riemannian manifolds with fibre $H$ and compact structure group $H$, because you could also write it as a Borel construction $G=G\times_HH\to G/H$. Here, you regard $G$ as a principal bundle over $G/H$ with fibre $H$. $G/H$ is equipped with the induced Riemannian metric, and $T^HG\subset TG$ is defined as $\ker(dp)^\perp$. Together with the induced metric on $H$, the Borel construction gives back the original metric $g$. But then its fibres are automatically totally geodesic, so now the answer is "yes".
Old answer to a different question :-)
Consider Berger metrics on $SU(2)\cong S^3\subset\mathbb C^2$.
Here the round metric is stretched or shrunk along the fibres of the Hopf
fibration $S^3\to S^2$. These metrics are still left invariant, but
not biinvariant. For $H$, take a one parameter subgroup, so $H\cong S^1$.
The induced metric on $H$ is still left invariant, hence biinvariant, because $H$ is abelian.
If all one parameter subgroups were geodesics, then their left translates by elements of $SU(2)$ would be geodesics, too. So the Berger metrics would have the same geodesics as the round metric, but that is not the case. Hence, the answer in general is no.
