Are one-dimensional tensors of arbitrary rank just scalars? Consider a tensor of arbitrary rank (2 for this case) $A_{ij}$, and dimension one. Granted there are two indices to specify a component, but since each index can only take one value, there is only one component in this entire tensor: $A_{11}$. So, are all one dimensional tensors scalars?
Further. transformation under coordinate transform for this case:
$$(A')^{11}={\left (\frac{\partial x'}{\partial x}\right )}^2A^{11}$$
suggests that since in general $(A')^{11}$ is not equal to $A^{11}$, it is not a scalar.
So what exactly is this non-scalar one component object?
 A: Perhaps an example is in order. Consider e.g. a 1D charge density $\rho$ in a 1D world. It transforms as a covariant (0,1) tensor $\rho^{\prime}=\frac{\partial x}{\partial x^{\prime}}\rho,$ so it is not a scalar.
A: Tensors of rank $k$ over a vector space $V$  form the space $\otimes^k V$. When the vector space is 1d, then we may as well take it to be the ground field $\mathbb{K}$.
But $\otimes^k \mathbb{K} \simeq \mathbb{K}$.
A tensor is an element of the left side, so equivalently it is also an element of the right side and so it is just a scalar.
So, yes, it's true. Alternatively:
Fix a 1d connected manifold without boundary $M$. Now, there are only two such manifolds: the circle and the infinite line. They are both orientated and flat. That they are orientated means that they have a volume form (this is a nowhere vanishing top form) and that they are flat means that they are isometric to the standard circle or Euclidean line. So we may as well choose this metric.
Since the manifold $M$ is 1d, the forms of rank 1 are exactly the top forms. There are no higher rank forms, they vanish by antisymmetry. Now choosing a top form, say:

$\omega \in \Omega^1M$

This is not a scalar form, however because of the flatness of 1d manifolds, there is a standard metric and we can use the Hodge star to convert to a scalar field:

$*\omega \in \Omega^0 M \simeq C^{\infty}M$

In brief:

$\Omega^1 M \simeq \Omega^0 M \simeq C^{\infty} M$.

This resolves the paradox of a 1d charge density that is not a scalar field as referred by @QMechanic in another post. That is not the full picture because it is also correct that it is canonically a scalar field.
However, differential forms are not the only tensors on a 1d manifold, they are only the antisymmetric covariant tensors. However, the argument in the first paragraph shows that they are also isomorphic to scalar fields.
