= Solution
\b[True.] Given a compatible <left-invariant group metric> $d$, define
$$
\boxed{d_R(x,y)=d(x^{-1},y^{-1}).}
$$
Pullback by the bijection $x\mapsto x^{-1}$ preserves all metric axioms, and because inversion is a <homeomorphism>, $d_R$ induces the same topology. For right translation by $a$, left invariance gives
$$
d_R(xa,ya)=d(a^{-1}x^{-1},a^{-1}y^{-1})=d(x^{-1},y^{-1})=d_R(x,y).
$$
Thus $d_R$ is the required compatible <right-invariant group metric>. No metrizability theorem is needed.
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