Let $K/\mathbb{Q}$ be a Galois number field and let $K_{\infty}/K$ be its cyclotomic $\mathbb{Z}_p$-extension defined by $K_{\infty}=K\cdot \mathbb{Q}_{\infty}$, where $\mathbb{Q}_{\infty}$ is the unique (cyclotomic) $\mathbb{Z}_p$-extension of $\mathbb{Q}$.
Let $\tau \in \text{Gal}(K_{\infty}/K)$ be a topological generator for $\text{Gal}(K_{\infty}/K)$ and for each $g \in \text{Gal}(K/\mathbb{Q})$, we fix a lift $\tilde{g} \in \text{Gal}(K_{\infty}/\mathbb{Q})$ of $g$.
I would like to know under which conditions one has that $\tilde{g}\tau=\tau\tilde{g}$, for all $g \in \text{Gal}(K/\mathbb{Q})$. This seems to be true if one assumes that $K\cap \mathbb{Q}_{\infty}=\mathbb{Q}$, but is it possible to find a necessary and sufficient condition?
I'm not sure there's much one can say in general. The condition that $\Bbb Q_\infty$ and $K$ be linearly disjoint over $\Bbb Q$ gives a direct product representation which of course implies the commutator relation you want. However, if we take the fully general approach, we see
This of course is how we know $\text{Gal}(K_\infty / K)$ is isomorphic to a subgroup of $\widehat{\Bbb Z}$. But the usual s.e.s. is
and a lift of $g\in \text{Gal}(K/\Bbb Q)$ is just any element of $\stackrel{\sim}{g}\text{Gal}(K_\infty/K)$. And the condition any such element commutes with your $\tau$ is just that $\text{Gal}(K_\infty/K)\le Z( \text{Gal}(K_\infty/\Bbb Q))$ since all elements of $\text{Gal}(K_\infty/\Bbb Q)$ are in some coset of some $\stackrel{\sim}{g}$. And as such this condition is both necessary and sufficient.
This of course recovers the case you mention, as then
and the right factor is abelian.