Does a scalar represent a linear map?

Any linear map from a one dimensional vector space to another is given by multiplication by a constant, or a $1\times 1$ matrix, once you've chosen basis (bases).

The $1\times 1$ matrix however changes depending on the choice of basis. A scalar in linear algebra is usually and element of the field the vector space is over, which is independent of the choice of basis for the vector space itself.

I'll expand on snulty's answer by giving an example.

Let's take a linear map $f:\Bbb R\to \Bbb R$ defined by $f(x) = 3x$. First I choose the standard basis, $\{1\}$, for both the domain and codomain of this map. Then clearly the matrix which represents this map is $(3)$.

But let's try a different pair of bases. I'll choose the basis $\{2\}$ for the domain and $\{3\}$ for the codomain of this function. Then the matrix which represents this map is $\left(2\right)$. How do I know? Well let's check what happens when we try to transform the "vector" $6$. The component of $6$ wrt the basis $\{2\}$ is $(3)$ so we multiply out the matrices $$\left(2\right)(3) = \left(6\right)$$ which is the component of the vector $18$ wrt the basis $\{3\}$.

So as you can see the $3$ in the definition of the map is a member of the field $\Bbb R$ and is independent of the bases chosen for the domain and codomain of the linear function. However, the sole entry of the matrix which represents the function depends on the bases we choose for the domain and codomain of our mapping.

If $A$ is an $n \times m$ matrix representing some linear map from $\mathbb{R}^m$ to $\mathbb{R}^n$, then it is helpful to note that the $j$th column of $A$ is the image of the $j$th basis vector (e.g., if you are using the standard basis, then it is the vector with a $1$ in the $j$th component, and all other components zero). Sticking with the standard basis, the columns of $A$ are elements of the image space $\mathbb{R}^n$.

If $A=[r]$ is $1 \times 1$, then yes it is a linear map $\mathbb{R} \to \mathbb{R}$. Under the standard basis, this does take the form $x \mapsto rx$ as you noted. From the previous paragraph, you can also think of the single entry $r$ as the image of the element $1$ (since $r \cdot 1 = r$), and clearly $r$ is an element of the image space $\mathbb{R}$.

I am not sure what your confusion is concerning scalars. The entries of vectors and matrices are scalars. In particular, a $1$-dimensional vector and a $1\times 1$ matrix can be thought of as a scalar, but can also still be interpreted as a vector or a matrix / linear map respectively.