First of all, I'll constrain the discussion assuming:
1) Finite-dimensional vector spaces
2) Real Vector spaces
3) Talking just about contravariant tensors
4) Physics which use the standard notion of Spacetime
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To answer your question I need to talk a little bit about Tensors.
I) The tensor object and pure mathematics:
The precise answer to the question "What is a tensor?" is, by far:
A tensor is a object of a vector space called Tensor Product.
In order to this general statement become something that have some value to you, I would like you to think a little bit about vectors and their algebra : the linear algebra.
I.1) What truly is a Vector?
First of all, if you look on linear algebra texts, you'll rapidly realize that the answer to the question "What are vectors after all? Matrices? Arrows? Functions?" is:
A vector is a element of a algebraic structure called vector space.
So after the study of the definition of a vector space you can talk with all rigour in the world that a vector isn't a arrow or a matrix, but a element of a vector space.
I.1.1) Some facts about vectors
Consider then a vector formed by a linear combination of basis vectors:
$$\mathbf{v} = \sum_{j = 1}^{n} v^{j}\mathbf{e}_{j} \tag{1}$$
This is well-known fact about vectors. So, there's another key point about basis vectors: the vector space is spanned by these basis vectors. You can create a "constrain machinery" to verify if a set of vectors spans a entire vector space (i.e. forms a basis):
A set $\mathcal{S}$ is a basis for a vector space $\mathfrak{V}$ if:
1) the vectors of the set $\mathcal{S}$ are linear independent
2) the vectors of the set $\mathcal{S}$ spanned the vector space $\mathfrak{V}$, i.e. $\mathfrak{V} \equiv span(\mathcal{S})$
So another point of view to "form" an entire vector space is from basis vectors. The intuitive idea is that, more or less, like if the basis vectors "constructs" (they span) and entire vector space.
Another fact is that you can change the basis $\mathbf{e}_{j}$ to another set basis of basis $\mathbf{e'}_{j}$. Well, when you do this the vector components suffer a change too. And then the components transforms like:
$$v'^{k} = \sum^{n}_{j=1}M^{k}_{j}v^{j}\tag{2}$$
but, of course, the vector object, remains the same:
$$\mathbf{v} = \sum_{j = 1}^{n} v^{j}\mathbf{e}_{j} = \sum_{j = 1}^{n} v'^{j}\mathbf{e'}_{j}$$
So, a vector, truly, is a object of a vector space, which have the form of $(1)$ and their components transforms like $(2)$.
I.1.2) The "physicist way" of definition of a Tensor
When you're searching about tensors on physics/engeneering texts you certainly will encounter the following definition of a tensor:
A Tensor is defined as the kind of object which transforms, under a coordinate transformation, like:
$$T'^{ij} = \sum^{n}_{k=1}\sum^{n}_{l=1} M^{i}_{k}M^{j}_{l} T^{kl} \tag{3}$$
This definition serves to encode the notion that a valid physical law must be independent of coordinate systems (or all that G.Smith said).
Well, there's some interesting happening here. A vector, is a object which have a precise formulation in terms of a algebraic structure, have a precise form (that of $(1)$, which the basis vectors spans the entire $\mathfrak{V}$) and their components have "transformation behaviour" like $(2)$. If you compare what I exposed about vector and $(3)$, you may reach the conclusion that, concerning about tensors, some information about their nature is missing.
The fact is, the definition $(3)$ isn't a tensor, but the "transformation behaviour" of the components of a tensor $\mathbf{T}$.
I.2) What truly is a Tensor?
Well, you have the transformation of components of a tensor, i.e. $(3)$, well defined. But what about their "space" and "form" (something like $(1)$)?
So, the space is called tensor product of two vector spaces:
$$V\otimes W \tag{4}$$
The construction of tensor product is something beyond the scope of this answer [*]. But the mathematical considerations about tensor products are that they generalize the concept of products of vectors (remember that, in linear algebra and analytic geometry you're able to "multiply" vector just using the inner product and vector product), they construct a concept of products of vector spaces (remember that a Direct sum of vector space gives you a notion of Sum of vector spaces), and they construct the precise notion of a tensor. Also, by the technology of the construction of the tensor product we can identify (i.e. stablish a isomorphism between vector spaces) the vector space $V\otimes W$ and $\mathfrak{Lin}^{2}(V^{*} \times W^{*}; \mathbb{K})$:
$$V\otimes W \cong \mathfrak{Lin}^{2}(V^{*} \times W^{*}; \mathbb{K}) \tag{5}$$
where $\mathfrak{Lin}^{2}(V^{*} \times W^{*}; \mathbb{K})$ is the dual vector space of all bilinear functionals.
So a tensor have the form:
$$\mathbf{T} = \sum^{n}_{i=1}\sum^{n}_{j=1} T^{ij} (\mathbf{e}_{i}\otimes\mathbf{e}_{j}) \tag{6}$$
And $\mathbf{T} \in V\otimes W$.
Well, given the transformation rule $(3)$ the space, $(5)$, and form, $(6)$, you can talk precisely about what tensor really is. It's clear that the "object tensor" isn't just a transformation of coordinates. Also, in $(6)$ the tensor basis $(\mathbf{e}_{i}\otimes\mathbf{e}_{j})$ spans $V\otimes W$.
By virtue of the general construction of tensor product and the identification given by $(5)$, you'll also encounter the definition of a tensor as a multilinear object which spits scalars:
$$\begin{array}{rl}
\mathbf{T} :V^{*}\times W^{*} &\to \mathbb{K} \\
(\mathbf{v},\mathbf{w})&\mapsto \mathbf{T}(\mathbf{v},\mathbf{w})=: v^{i}\cdot_{\mathbb{K}}w^{j}
\end{array}$$
Where the operation $\cdot_{\mathbb{K}}$ is the product defined in the field.
With this picture we say that a tensor like $(6)$ is a tensor of rank 2. And a vector a tensor of rank 1. Furthermore a scalar a tensor of rank 0.
II) The tensor object and physics
The well stablish physics, in general, deals with spacetime (like Newtonian physics), and the theory of spacetime is geometry. So, in order to really apply the tensor theory in physics first we have to give the geometry of physics.
The geometry is basically classical Manifold Theory (which, again, is beyond the scope). And by Manifold Theory we can apply tensors on Manifolds introducing the concept of a tangent vector space. In parallel, we can construct another algebraic structure called Fibre Bundle of tangent spaces and then create the precise notion of Vector Field and Tensor Field.
Tensor Fields are the real objects defined in physics books as tensors and we use the word of a tensor and tensor field as synonyms (IN FACT THEY ARE NOT THE SAME CONCEPT!). A tensor field is a section of the tensor bundle and a vector field, a section of vector bundle. But the intuitive definition (by far, general to physics) of a tensor field is then:
$$[\mathbf{T}(x^{k})] = \sum^{n}_{i=1}\sum^{n}_{j=1} [T^{ij}(x^{k})] ([\mathbf{e}_{i}(x^{k})]\otimes[\mathbf{e}_{j}(x^{k})]) \tag{5}$$
A tensor field is the object which attaches a tensor to every point p of the Manifold.
With the manifold theory, the transformation rule becomes:
$$[T'^{ij}(x^{m})] = \sum^{n}_{k=1}\sum^{n}_{l=1} \frac{\partial x'^{i}}{\partial x^{k}}\frac{\partial x'^{j}}{\partial x^{l}} [T^{kl}(x^{m})] \equiv T'^{ij} = \sum^{n}_{k=1}\sum^{n}_{l=1} \frac{\partial x'^{i}}{\partial x^{k}}\frac{\partial x'^{j}}{\partial x^{l}} T^{kl} \tag{7}$$
Notice that the partials are simply the transformation matrices $M$. The matrices $M$ are called the Jacobians transformation matrices and the matrices $M$ became these jacobians by virtue of Manifold theory.
In a restric way, these Jacobians are rotations,lorentz transformations,galilean transformation, and so on.
III) What is the difference between zero-rank tensor x (scalar) and 1D vector [x]?
So, in order to talk about lengths we have to realize that we are talking about a scalar field, or a tensor of rank 0 (i.e. we've stablish an isomorphism and then we can say that a scalar field is isomorphic to a rank zero tensor). Then the difference between a scalar and a 1D vector (which is isomorphic to a tensor of rank 1) is that one is a scalar field and the other is a vector field. From a "Pure" mathematical point of view, (section I) if this answer) one is a member of the field $\mathbb{K}$ and the other is a member of a vector space.
Also, you're quite right, a scalar (or a scalar field) is a rank 0 tensor or "a object which do not have "matrices of change"; an object which do not suffer a change under a transformation of coordinates (we say that a scalar quantity is a invariant quantity).
$$\phi'= \phi$$
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[*] ROMAN.S. Advanced Linear Algebra. Springer. chapter 14. 1 ed. 1992.
