$\newcommand{\inv}{\text{inv}}$
Let $G$ be a Lie group. Assume $d$ is a metric on $G$ (in the sense of metric spaces) which is bi-invariant. Is it true that the inverse automorphism must be an isometry of $(G,d)$?
I do not assume $d$ is induced by a Riemannian metric*.
*If $d$ is induced by a Riemannian metric $g$, than the answer is positive:
By the Myers–Steenrod theorem, $g$ is also bi-invariant.
Note that $\inv = R_{s^{-1}}\circ \inv \circ L_{s^{-1}}$, so $$ (d\inv)_s = (dR_{s^{-1}})_e \circ (d\inv)_e \circ (dL_{s^{-1}})_s $$
Since $(d\inv)_e:T_eG \to T_eG$ is the minus operation $(v \mapsto -v)$, we get:
$$ (d\inv)_s = - (dR_{s^{-1}})_e \circ (dL_{s^{-1}})_s $$
So, bi-invariance of the metric $g$ implies inverse-invariance of $g$, which implies invers-invariance of $d$.
I understand the bi-invariance as $$ d(ax,ay)=d(x,y)=d(xa,ya) $$ for any $a,x,y\in G$. Then $$ d(x,y)=d(1,x^{-1}y)=d(y^{-1},x^{-1})=d(x^{-1},y^{-1}). $$ The last step is the symmetry of $d$.