For a vector field $A_\mu$, there are infinitely many configurations that describe the same physical situation. This is a result of our gauge freedom $$ A_\mu (x_\mu) \to A'_\mu \equiv A_\mu (x_\mu) + \partial_\mu \eta(x_\mu ),$$ where $\eta (x_\mu)$ is an arbitrary scalar function.
Therefore, each physical situation can be described by an equivalence class of configurations. All members within a given equivalence class are related by a gauge transformation. Configurations in different equivalence classes describe physically distinct situations and therefore are not related by gauge transformations.
To fix the gauge, we need to pick exactly one member from each such equivalence class. A popular way to accomplish this is by demanding \begin{equation} \partial_\mu A^\mu =0 \, . \end{equation} Apparently this works because there is only exactly one member in each equivalence class that fulfills this additional condition. How can this be shown and understood?
Answer
The Lorenz gauge condition does not fix the gauge completely.
Let $A^\mu$ be a field satisfying the Lorenz gauge condition $\partial_\mu A^\mu = 0$. Given a scalar function $f$, let $B^\mu = A^\mu + \partial^\mu f$. $B^\mu$ can also satisfy the Lorenz gauge condition if
$$ \partial_\mu B^\mu = \partial_\mu\partial^\mu f = 0, $$ i.e. if $f$ is the Minkowski space equivalent of a harmonic function. Therefore it is not true that the condition picks exactly one function per gauge-equivalence class. In each class, there is a whole (non-trivial) vector space of functions satisfying the Lorenz gauge.
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