The QED Lagrangian (for one fermion) is:
$$\mathcal{L} = -\frac{1}{4}F_{\mu \nu}F^{\mu \nu} + \bar{\psi}\left(i \gamma^{\mu}\partial_{\mu} - m \right) \psi - q \bar{\psi} \gamma^{\mu}A_{\mu}\psi = \mathcal{L}_{kin} - J^{\mu}A_{\mu}$$
$\mathcal{L}$ has a global symmetry: $\psi \rightarrow e^{-i q} \psi$ and the associated Noether current is:
$$J^{\mu} = q \bar{\psi} \gamma^{\mu} \psi$$
Now we can consider the 3-point Green function with an insertion of one current $J_{\mu}$:
$$G^{(3)}_{\mu}=\langle \mathcal{T}[J_{\mu} \psi \bar{\psi}] \rangle$$
Because, if we call S the fermion propagator, using the Ward Identity in the momentum space we obtain:
$$-ik^{\mu}S^{-1}(p+k)G^{(3)}_{\mu}(p,k)S^{-1}(p) = q \left(S^{-1}(p)- S^{-1}(p+k)\right) \tag{1}$$
Now, if we call $D_{\mu \nu}$ the photon propagator, applying the LSZ formula we obtain that the Amputated Vertex $-iq\Gamma_{\mu} (p,q)$ is:
$$-iq\Gamma_{\mu}(p,k) = S^{-1}(p) S^{-1}(p+k) D^{-1}_{\mu \nu}(k) \langle \mathcal{T}[A^{\nu} \psi \bar{\psi}] \rangle$$
So far so good. What I don't understand is the following:
Why is it true that the las of the last Ward Identity (1) is equal to the Amputated Vertex $-iq\Gamma_{\mu} (p,k)$ contracted with $(-k^{\mu})$? Or in other words why it is true that: $$iS^{-1}(p+k)G^{(3)}_{\mu}(p,k)S^{-1}(p) = -iq\Gamma_{\mu}(p,k)$$
I understand that the LHS of this last equation is indeed an Amputated Vertex, but it should be the amputated vertex of $J_{\mu}\bar{\psi}{\psi}$ and not of $A_{\mu}\bar{\psi}{\psi}$.
I think that it has something to do with the fact that $\mathcal{L}_{int}=-J^{\mu}A_{\mu}$ and so we can kinda substitute $J_{\mu}$ with $(-A_{\mu})$, but I don't know how, I'm not convinced about this also because the last equation should stand in general also for the Scalare QED where $\mathcal{L}_{int}=-J^{\mu}A_{\mu} + q^{2}A_{\mu}A^{\mu}\phi^{\dagger}\phi \neq -J^{\mu}A_{\mu}$.
Edit:
Maybe the answer is found using a Reduction Formula (LSZ):
$$- \langle 0 | J_{\mu}|0 \rangle = \langle 0 |A_{\mu} \rangle = D^{-1}_{\mu \nu} \langle 0 |A^{\nu} |0 \rangle$$
But I'm not sure about this either.
