This question touches some very interesting conceptual questions in AdS/CFT that are often omitted in presentations of the holographic dictionary. So what is the usual story of the holographic dictionary (considering a free scalar first in the Euclidean setting for simplicity first)
- There is a bulk scalar field with an equation of motion $(\Box-m^2)\phi=0$, where $\Box$ is the d'Alembert operator in the given background spacetime.
- The scalar can be expanded near the boundary
$$\phi=z^{\Delta_-}(\phi_{0}+ \phi_{2}z^{2}+\ldots+\phi_{d} z^{d}\log z+\ldots)+z^{\Delta_+}(\psi_{d}+\psi_{d+2}z^{\Delta_++2}+\ldots)$$
where the $\log$-term only appears if the $\Delta_+$ and $\Delta_-$ differ by an integer. One gets $\Delta_\pm=\frac{d}{2}\pm\sqrt{\frac{d^2}{4}+m^2}$ and $\phi_0$ and $\psi_d$ are the only independent coefficients reflecting the fact that we have a second order equation of motion. All other coefficients are found upon solving the equations of motion perturbatively.
- The solution to the equation of motion is plugged back into the action and after dealing with regularization issues one obtains a functional which is a boundary integral. This boundary integral is interpreted as the generating functional of correlators in the dual CFT where $\phi_0$ is the source of a scalar operator $\mathcal{O}$, called the dual operator, i.e.
$$
\langle e^{-\int d^dx \phi_0\mathcal{O}}\rangle_{\text{CFT}}=e^{-S_{\text{on-shell}}\left[\phi|_{\partial}=\phi_0\right]}.
$$
Up to normalization one finds $\langle\mathcal{O}\rangle\propto\psi_d$.
- The map $x^\mu\to x'{}^\mu =\lambda x^\mu$, $z\to z'= \lambda z$ is an isometry of the AdS metric
$$
ds^2=\frac{1}{z^2}\left(\eta_{\mu\nu} dx^\mu dx^\nu+dz^2\right)
$$
under which the bulk scalar transforms as $\phi'(x')=\phi(x)$. On the boundary, z=0, this isometry becomes a dilatation (which is not an isometry of the Minkowski metric) and from the expansion of the scalar above one can see that the transformed scalar is $\phi_0{}'=\lambda^{-\Delta_-}\phi_0$. Because of the coupling $\int d^dx \phi_0\mathcal{O}$ we can therefore see that $\mathcal{O}$ has scaling dimension $\Delta_+=d-\Delta_-$.
To summarize: In the expansion of the scalar close to the boundary there are two independent modes. The leading mode is interpreted as source of the dual operator in the CFT and one finds that the one-point function of the operator is given by the subleading mode.
Now, why am I repeating this well known story? Because under some circumstances, there is an alternative to the described procedure, namely $\phi_0$ and $\psi_d$ can switch roles. This was first explored by Klebanov & Witten. They describe that there is a bound on the scaling dimensions of scalar operators in the field theory, which is given by $\Delta>\frac{d-2}{2}$ and is required by unitarity. Now, the formula $\Delta_\pm=\frac{d}{2}\pm\sqrt{\frac{d^2}{4}+m^2}$ requires $\Delta_+>\frac{d}{2}$ and $\Delta_-<\frac{d}{2}$, so $\Delta_+$ is always above the unitarity bound no matter what the mass $m$ is and have seen that $\Delta_+$ turned out to be the scaling dimension of $\mathcal{O}$ in the procedure above, so this is consistent. However, in a small range of the mass parameter $m$, above the Breitenlohner-Freedman bound $m^2>-\frac{d}{2}$, but with $m^2<-\frac{d}{2}+1$, also $\Delta_->\frac{d-2}{2}$! So one can just as well suppose that $\psi_d$ is supposed to be the source of a scalar operator, which then turns out to have scaling dimension $\Delta_->\frac{d-2}{2}$, i.e. above the unitarity bound.
Now, we start to make connection with the original question. If one considers a gauge field $A$ in the bulk instead of a scalar, which is massless since a mass term breaks gauge invariance, we will encounter the same ambiguity. We can either dial the leading mode as source and find that the subleading mode determines the one point function of the dual operator or the other way around. The dual operator of a gauge field is a conserved current $J^\mu$ whose $t$-component is the charge density of the associated conserved charge, i.e. the expansion of the $t$-component of the bulk gauge field will look like
$$
A^t=\mu(z^{\Delta_-}+\ldots)+\langle J^t\rangle(z^{\Delta_+}+\ldots).
$$
Because of the ambiguity of which of the two modes we choose to dial as a source, and which we want to read off as a response to this dialling, we can either force the chemical potential or the charge density to have a certain value, and therefore either consider the grand canonical or the canonical ensemble. Therefore, it is our free choice to work in either the canonical or the grand canonical ensemble in the field theory. The chemical potential and the charge density play the same role in both cases, it is just the question which one is fixed by hand and which is responding dynamically, such that we can compute its expectation value via the dictionary.