$\color{blue}{\textbf{ANSWER A}}\:$ (based on charge invariance, paragraph extracted from Landau)
The answer is given in ACuriousMind's comment as pointed out also by WetSavannaAnimal aka Rod Vance. Simply I give the details copying from "The Classical Theory of Fields", L.D.Landau and E.M.Lifshitz, Fourth Revised English Edition :
$\boldsymbol{\S}\: \textbf{28. The four-dimensional current vector}$
Instead of treating charges as points, for mathematical convenience we frequently consider them to be distributed continuously in space. Then we can introduce the "charge density" $\:\varrho\:$ such that $\:\varrho dV\:$ is the charge contained in the volume $\: dV$. The density $\:\varrho\:$ is in general a function of the coordinates and the time. The integral of $\:\varrho\:$ over a certain volume is the
charge contained in that volume.......
.......The charge on a particle is, from its very definition, an invariant quantity, that is, it does not depend on the choice of reference system. On the other hand, the density $\:\varrho\:$ is not generally an invariant--only the product $\:\varrho dV\:$ is invariant.
Multiplying the equality $\:de=\varrho dV\:$ on both sides with $\:dx^{i}\:$:
\begin{equation}
de\,dx^{i}=\varrho dVdx^{i}=\varrho dVdt\dfrac{dx^{i}}{dt}
\nonumber
\end{equation}
On the left stands a four-vector (since $\:de\:$ is a scalar and $\:dx^{i}\:$ is a four-vector). This means that the right side must be a four-vector. But $\: dVdt\:$ is a scalar(1), and so $\:\varrho dx^{i}/dt\:$ is a four-vector.This vector (we denote it by $\:j^{i}$) is called the current four-vector:
\begin{equation}
j^{i}=\varrho \dfrac{dx^{i}}{dt}.
\tag{28.2}
\end{equation}
The space components of this vector form the current density vector,
\begin{equation}
\mathbf{j}=\varrho \mathbf{v},
\tag{28.3}
\end{equation}
where $\:\mathbf{v}\:$ is the velocity of the charge at the given point. The time component of the four vector (28.2) is $\:c\varrho$. Thus
\begin{equation}
j^{i}=\left(c\varrho ,\mathbf{j}\right)
\tag{28.4}
\end{equation}
(1)
Note by Frobenius : We have
\begin{equation}
dVd(ct)=dx^{1}dx^{2}dx^{3}dx^{4}
\tag{01}
\end{equation}
Now, for the relation between the infinitesimal 4-volumes in Minkowski space
\begin{equation}
dx'^{1}dx'^{2}dx'^{3}dx'^{4} =\begin{vmatrix}
\dfrac{\partial x'_{1}}{\partial x_{1}}& \dfrac{\partial x'_{1}}{\partial x_{2}}&\dfrac{\partial x'_{1}}{\partial x_{3}}&\dfrac{\partial x'_{1}}{\partial x_{4}}\\
\dfrac{\partial x'_{2}}{\partial x_{1}}& \dfrac{\partial x'_{2}}{\partial x_{2}}&\dfrac{\partial x'_{2}}{\partial x_{3}}&\dfrac{\partial x'_{2}}{\partial x_{4}}\\
\dfrac{\partial x'_{3}}{\partial x_{1}}& \dfrac{\partial x'_{3}}{\partial x_{2}}&\dfrac{\partial x'_{3}}{\partial x_{3}}&\dfrac{\partial x'_{3}}{\partial x_{4}}\\
\dfrac{\partial x'_{4}}{\partial x_{1}}& \dfrac{\partial x'_{4}}{\partial x_{2}}&\dfrac{\partial x'_{4}}{\partial x_{3}}&\dfrac{\partial x'_{4}}{\partial x_{4}} \end{vmatrix}
dx^{1}dx^{2}dx^{3}dx^{4}=\left\vert\dfrac{\partial\left(x'^{1},x'^{2},x'^{3},x'^{4}\right)}{\partial\left(x^{1},x^{2},x^{3},x^{4}\right)}\right\vert dx^{1}dx^{2}dx^{3}dx^{4}
\tag{02}
\end{equation}
where $\:\left\vert\partial\left(x'^{1},x'^{2},x'^{3},x'^{4}\right)/\partial\left(x^{1},x^{2},x^{3},x^{4}\right)\right\vert\:$ the Jacobian, that is determinant of the Jacobi matrix. But the Jacobi matrix is the Lorentz matrix $\:\Lambda\:$ with $\:\det(\Lambda)=+1$,
that is
\begin{equation}
\left\vert\dfrac{\partial\left(x'^{1},x'^{2},x'^{3},x'^{4}\right)}{\partial\left(x^{1},x^{2},x^{3},x^{4}\right)}\right\vert=\det(\Lambda)=+1
\tag{03}
\end{equation}
so
\begin{equation}
dx'^{1}dx'^{2}dx'^{3}dx'^{4} =dx^{1}dx^{2}dx^{3}dx^{4}=\text{scalar invariant}
\tag{04}
\end{equation}
Best Answer
This works perfectly fine, but is more a heuristic.
If $\lambda$ is a scalar function, meaning that $\lambda \mapsto \lambda' = \lambda\circ \Lambda^{-1}$ under a Lorentz transformation $\Lambda$, then $\lambda(x) \mapsto \lambda'(x') = \lambda(\Lambda^{-1}\Lambda x) = \lambda(x)$. Most functions that look as if they are scalars are scalars, exceptions usually involve derivatives in some form. In particular $f(x) = x^0$ is a perfectly fine scalar function, even though it is not a Lorentz scalar in the sense that $f(\Lambda x) = f(x)$. It's confusing terminology.
In the same vein, transforming as a four-vector under Lorentz transformations means $A^\mu \mapsto \Lambda_\nu^\mu A'^\nu$ (note again the prime, standing for $A' = A\circ\Lambda^{-1}$), or $A\mapsto \Lambda\circ A\circ \Lambda^{-1}$, since then $A^\mu(x) \mapsto \Lambda^\mu_\nu A'^\nu(x') = \Lambda^\mu_\nu A^\nu(x)$ - the point at which you evaluate the function after the transformation still hasn't changed, but the transformation not only changed the way the coordinate is expressed (as $x'$ instead of $x$) but also the basis of your vector space.
So, finally, yes $\partial^\mu \lambda$ is a four-vector if $\lambda$ is a scalar function simply because $\partial^\mu \mapsto \Lambda^\mu_\nu \partial'^\nu$ and $\lambda(x)\mapsto \lambda'(x')$. Since adding two things that are not of the same type is generally not very well-defined, we conclude that $A^\mu$ better be a four-vector if it is to be a meaningful quantity. However, "better be" is not a proof. Formally you have to examine your definition of $A^\mu$ and deduce from that that it is a four-vector. How exactly that works depends on whether you cobbled it together from the non-relativistic parts $\phi,\vec A$ or defined it to be the anti-derivative of the field-strength tensor $F$.