Well, I played around a bit with something I thought might make a counterexample, but I couldn't make it work (or rather, it always did work). Maybe something "more non-commutative" (?) than $\mathrm{Gal}(\sqrt[n]{a},\zeta_m)$ needs to be used; or maybe the statement is true. I'll record my efforts here as a CW answer in case it gives someone else an idea.
Let $K=\mathbb{Q}(\sqrt[3]{2},\zeta_{15})$, and let $\phi,\rho\in\mathrm{Gal}(K/\mathbb{Q})$ be the automorphisms defined by $\phi(\sqrt[3]{2})=\zeta_{15}^5\sqrt[3]{2},\quad \phi(\zeta_{15})=\zeta_{15}$ $\rho(\sqrt[3]{2})=\sqrt[3]{2},\quad \rho(\zeta_{15})=\zeta_{15}^4$ The automorphism $\rho$ certainly can be extended to $\overline{\mathbb{Q}}$; let $\psi\in\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ be any such extension. Then $\phi$ and $\psi|_K=\rho$ commute because $\phi(\rho(\sqrt[3]{2}))=\phi(\sqrt[3]{2})=\zeta_{15}^5\sqrt[3]{2}=\zeta_{15}^{20}\sqrt[3]{2}=\rho(\zeta_{15}^5\sqrt[3]{2})=\rho(\phi(\sqrt[3]{2}))$ $\phi(\rho(\zeta_{15}))=\phi(\zeta_{15}^4)=\zeta_{15}^4=\rho(\zeta_{15})=\rho(\phi(\zeta_{15}))$
Let $L=\mathbb{Q}(\sqrt[15]{2},\zeta_{15})$. Any extension $\sigma\in\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ of $\phi$ to an automorphism of $\overline{\mathbb{Q}}$ must restrict down to $L$ to do one of the following 5 things: $\sigma|_L(\sqrt[15]{2})\in\{\zeta_{15}\sqrt[15]{2},\;\zeta_{15}^4\sqrt[15]{2},\;\zeta_{15}^7\sqrt[15]{2},\;\zeta_{15}^{10}\sqrt[15]{2},\;\zeta_{15}^{13}\sqrt[15]{2}\},\quad \sigma|_L(\zeta_{15})=\zeta_{15}$ Now suppose we've taken our $\psi\in\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ such that $\psi|_L(\sqrt[15]{2})=\sqrt[15]{2}$ and $\psi|_L(\zeta_{15})=\zeta_{15}^4$ (e.g., take the automorphism of $L$ that does that, and lift it to $\overline{\mathbb{Q}}$). Then $\psi|_L$ and $\sigma|_L$ can never commute, because $\sigma|_L(\psi|_L(\sqrt[15]{2}))=\sigma|_L(\sqrt[15]{2})=\zeta_{15}^{1,4,7,10,13}\sqrt[15]{2}\neq\zeta_{15}^{4,1,13,10,7}\sqrt[15]{2}=\psi|_L(\sigma|_L(\sqrt[15]{2}))$ (except, of course, this is wrong; if $\sigma|_L(\sqrt[15]{2})=\zeta_{15}^{10}\sqrt[15]{2}$ then we get the same answer on both sides. This is also the case if we'd started with $\rho(\zeta_{15})=\zeta_{15}^{13}$, which is the only other possible choice for this particular setup.)