$\newcommand\F{\mathbf{F}}$ $\newcommand\Z{\mathbf{Z}}$ $\newcommand\Q{\mathbf{Q}}$ $\newcommand\Gal{\mathrm{Gal}}$

This answer is basically a response to Speyer (although he omitted this part of the argument, so hopefully it is correct). edit: This doesn't seem to work.

Suppose that $G = \Gal(K/\Q)$, where $K$ is unramified over $\Q(\zeta_p)$. Let $V = \Gal(\Q(\zeta_p)/\Q)$. As noted by Speyer, byconsidering the inertia group, we have that that $G$ is a semi-direct product, so $G = (\Z/p \Z)^{\times} \ltimes V$, and elements of $G$ have the form $(c,g):=(c,0)(0,g)$. Since the genus field of $\Q(\zeta_p)$ is trivial, we certainly have $V = [G,G]$. Let us now assume that

$$H = [G,[G,G]] = [G,V] \ne V.$$

Then we can define a function $w: G \rightarrow \Z/p$ on $G$ as follows:

$$\psi(c,g) = \begin{cases} c, & g \in H, \\ 0, & g \notin H. \end{cases}$$

For fixed $x$ and $y$ in $V$,

$$\psi((0,x)(c,0)(0,y^{-1})) = \psi(c,c^{-1} xc y^{-1}) = \psi(c,[c^{-1},x] x y^{-1}).$$

Now $[c^{-1},x] \in [G,V] = H$, and hence

$$\psi((0,x)(c,0)(0,y^{-1})) = \begin{cases} c, & xH = yH, \\ 0, & \text{otherwise}.\end{cases}$$

If I understand correctly, this satisfies the properties of $w$ desired by Speyer.

edit:I guess this doesn't work --- or rather not in an interesting way --- the conditions are never satisfied! They force $G/H$ to be abelian and then $G/H = G/V$.

$\newcommand\F{\mathbf{F}}$ $\newcommand\Z{\mathbf{Z}}$ $\newcommand\Q{\mathbf{Q}}$ $\newcommand\Gal{\mathrm{Gal}}$ $\newcommand\GL{\mathrm{GL}}$ $\newcommand\SL{\mathrm{SL}}$

Some more examples of groups with corresponding maps $w$:

Let $G = \GL_2(\F_p)$ and $V = \SL_2(\F_p)$, and consider $(\Z/p)^{\times} \subset G$ via the map

$$c \rightarrow \left( \begin{matrix} c & 0 \\ 0 & 1 \end{matrix} \right).$$

Then one can take $w: G \rightarrow \F_p$ to be (say) the $[1,1]$ entry of $G$, because then

$$w \left( \left( \begin{matrix} w & x \\ y & z \end{matrix} \right) \left( \begin{matrix} c & 0 \\ 0 & 1 \end{matrix} \right) \left( \begin{matrix} r & s \\ t & u \end{matrix} \right)\right) = c p w + r x$$

is affine linear in $c$. Hence this gives an admissible $w$. Moreover, for $G$ to occur as a Galois group, we would want a representation:

$$\rho: \Gal(\overline{\Q}/\Q) \rightarrow \GL_2(\F_p)$$

which is unramified outside $p$ with cyclotomic determinant and so that the restriction to inertia was

$$\rho |_{I_p} = \left( \begin{matrix} \varepsilon & 0 \\ 0 & 1 \end{matrix} \right)$$

where $\varepsilon$ is the mod-$p$ cyclotomic character (which gives the canonical identification of $\Gal(\Q_p(\zeta_p)/\Q_p)$ with $(\Z/p)^{\times}$).

Speyer asked in another question whether one could prove that $V_{p-2} = 0$ using anything simpler than Herbrand --- perhaps with the idea that any simple direct negative answer to this question would give a new proof. This example is even worse --- to rule out the existence of such a representation $\rho$, I can't see any simpler argument than using the proof of Serre's conjecture by Khare-Wintenberger (the case $N(\rho) = 1$ and $k(\rho) = 2$). Even worse, if one replaces $\GL_2$ with $\GL_3$ and maps $(\Z/p)^{\times}$ to $G$ via $\mathrm{diag}(c,1,1)$, it becomes an open problem to show that a corresponding extension with Galois group $\GL_3(\F_p)$ and representation $\rho$ does not exist --- although various standard conjectures certainly imply that it does not. This strongly suggests that proving the answer to the original question is "no" will be very hard. So either one should

1. Try to prove the answer is "conditionally no" by using conjectures in number theory, after more fully understanding the group theoretic condition.

2. Try to prove the answer is "yes".

earlier: This answer is basically a response to Speyer (although he omitted this part of the argument, so hopefully it is correct). edit: This doesn't seem to work.

Suppose that $G = \Gal(K/\Q)$, where $K$ is unramified over $\Q(\zeta_p)$. Let $V = \Gal(\Q(\zeta_p)/\Q)$. As noted by Speyer, byconsidering the inertia group, we have that that $G$ is a semi-direct product, so $G = (\Z/p \Z)^{\times} \ltimes V$, and elements of $G$ have the form $(c,g):=(c,0)(0,g)$. Since the genus field of $\Q(\zeta_p)$ is trivial, we certainly have $V = [G,G]$. Let us now assume that

$$H = [G,[G,G]] = [G,V] \ne V.$$

Then we can define a function $w: G \rightarrow \Z/p$ on $G$ as follows:

$$\psi(c,g) = \begin{cases} c, & g \in H, \\ 0, & g \notin H. \end{cases}$$

For fixed $x$ and $y$ in $V$,

$$\psi((0,x)(c,0)(0,y^{-1})) = \psi(c,c^{-1} xc y^{-1}) = \psi(c,[c^{-1},x] x y^{-1}).$$

Now $[c^{-1},x] \in [G,V] = H$, and hence

$$\psi((0,x)(c,0)(0,y^{-1})) = \begin{cases} c, & xH = yH, \\ 0, & \text{otherwise}.\end{cases}$$

If I understand correctly, this satisfies the properties of $w$ desired by Speyer.

edit:I guess this doesn't work --- or rather not in an interesting way --- the conditions are never satisfied! They force $G/H$ to be abelian and then $G/H = G/V$.

$\newcommand\F{\mathbf{F}}$ $\newcommand\Z{\mathbf{Z}}$ $\newcommand\Q{\mathbf{Q}}$ $\newcommand\Gal{\mathrm{Gal}}$

This answer is basically a response to Speyer (although he omitted this part of the argument, so hopefully it is correct). I don't usually give incomplete answers, but the enticement of (some share) of liquor proved irresistible.

Suppose that $G = \Gal(K/\Q)$, where $K$ is unramified over $\Q(\zeta_p)$. Let $V = \Gal(\Q(\zeta_p)/\Q)$. As noted by Speyer, byconsidering the inertia group, we have that that $G$ is a semi-direct product, so $G = (\Z/p \Z)^{\times} \ltimes V$, and elements of $G$ have the form $(c,g):=(c,0)(0,g)$. Since the genus field of $\Q(\zeta_p)$ is trivial, we certainly have $V = [G,G]$. Let us now assume that

$$H = [G,[G,G]] = [G,V] \ne V.$$

Then we can define a function $w: G \rightarrow \Z/p$ on $G$ as follows:

$$\psi(c,g) = \begin{cases} c, & g \in H, \\ 0, & g \notin H. \end{cases}$$

For fixed $x$ and $y$ in $V$,

$$\psi((0,x)(c,0)(0,y^{-1})) = \psi(c,c^{-1} xc y^{-1}) = \psi(c,[c^{-1},x] x y^{-1}).$$

Now $[c^{-1},x] \in [G,V] = H$, and hence

$$\psi((0,x)(c,0)(0,y^{-1})) = \begin{cases} c, & xH = yH, \\ 0, & \text{otherwise}.\end{cases}$$

If I understand correctly, this satisfies the properties of $w$ desired by Speyer. Of course, we then need to show that there exists a $K$ with the required properties. Perhaps the simplest construction is as follows. Let

$$\Gamma = \left( \begin{matrix} 1 & * & * \\ 0 & * & * \\ 0 & 0 & 1 \end{matrix} \right) \subset \mathrm{GL}_2(\mathbf{F}_3)$$

This is a group of order $54$, whose determinant gives a map to $\F^{\times}_3 \simeq \Z/2\Z$. The group $(\Z/p \Z)^{\times}$ also has a unique map (the quadratic residue) to this group. So let $G \subset \Gamma \times (\Z/p \Z)^{\times}$ be the index two subgroup consisting of pairs mapping to the same element in in $\Z/2 \Z$ under this map.

The group $V$ will be the unipotent subgroup of $\Gamma$, which is the unique non-abelian group of order $27$ all of whose elements have order $3$, and $H$ will be the order $3$ subgroup consisting of unipotent matrices in which just the upper right most corner is non-zero.

To construct such a Group as a Galois group (unramified over $\Q(\zeta_p)$) we want to construct a Galois extension of $E = \Q(\sqrt{p^*}) \subset \Q(\zeta_p)$ unramified at all finite primes with Galois group $\Gamma$. So one should look for an $E$ with class group containing $(\Z/3\Z)^2$. Now I could write a long screed about the corresponding extension problem and showing how to make it vanish, or I could just note that for the subgroup

$$\Phi = \left( \begin{matrix} 1 & * & 0 \\ 0 & * & 0 \\ 0 & 0 & 1 \end{matrix} \right) $$

of index $9$, we get an action of $\Gamma \rightarrow S_9$ coming from left cosets on $\Gamma/\Phi$, and the element of order $2$ acts with cycle shape $(**)(**)(**)$. Hence, given a field $E$ with the required class group, we want a number field of degree $9$ with discriminant $(p^*)^3$ with the right splitting field. The first three primes with the required class group are $p^* = -2437$, $-3299$, and $-4027$. And now we simply look up the tables here: https://hobbes.la.asu.edu/NFDB/ to find the polynomial

$$x^9 - 3x^8 + 11x^6 - 28x^5 + 39x^4 - 36x^3 + 27x^2 - 13x + 3$$

cuts out a number field of discriminant $-3299^3$ and Galois group $\Gamma$ (one can now easily check the splitting field is unramified and indeed as it has to be totally split over the quadratic extension) and so gives rise to the required unramified extension of $\Q(\sqrt{-3299})$ with Galois group $\Gamma$, and thus the compositum of this with $\Q(\zeta_{3299})$ gives the required Galois extension $K/\Q$ with Galois group $G$.

$\newcommand\F{\mathbf{F}}$ $\newcommand\Z{\mathbf{Z}}$ $\newcommand\Q{\mathbf{Q}}$ $\newcommand\Gal{\mathrm{Gal}}$

This answer is basically a response to Speyer (although he omitted this part of the argument, so hopefully it is correct). edit: This doesn't seem to work.

Suppose that $G = \Gal(K/\Q)$, where $K$ is unramified over $\Q(\zeta_p)$. Let $V = \Gal(\Q(\zeta_p)/\Q)$. As noted by Speyer, byconsidering the inertia group, we have that that $G$ is a semi-direct product, so $G = (\Z/p \Z)^{\times} \ltimes V$, and elements of $G$ have the form $(c,g):=(c,0)(0,g)$. Since the genus field of $\Q(\zeta_p)$ is trivial, we certainly have $V = [G,G]$. Let us now assume that

$$H = [G,[G,G]] = [G,V] \ne V.$$

Then we can define a function $w: G \rightarrow \Z/p$ on $G$ as follows:

$$\psi(c,g) = \begin{cases} c, & g \in H, \\ 0, & g \notin H. \end{cases}$$

For fixed $x$ and $y$ in $V$,

$$\psi((0,x)(c,0)(0,y^{-1})) = \psi(c,c^{-1} xc y^{-1}) = \psi(c,[c^{-1},x] x y^{-1}).$$

Now $[c^{-1},x] \in [G,V] = H$, and hence

$$\psi((0,x)(c,0)(0,y^{-1})) = \begin{cases} c, & xH = yH, \\ 0, & \text{otherwise}.\end{cases}$$

If I understand correctly, this satisfies the properties of $w$ desired by Speyer.

edit:I guess this doesn't work --- or rather not in an interesting way --- the conditions are never satisfied! They force $G/H$ to be abelian and then $G/H = G/V$.