What is the invariant in tensors? It is being said that the most important property of tensors is that they are invariant under basis changes.  
What exactly does stay invariant in tensors?  
Can you please explain this invariance for scalars, vectors, linear maps and metric tensors?
I know about this answer, but does this mean that the invariant for a tensor is that if the basis changes its component changes in a way so that it return the same number when applied to the same object?  
And if so then does this mean that things like a linear map matrix which accept a vector and return another vector are not tensors because they are not returning a scalar?
 A: An example of this phenomenon can be illustrated with a constructions similar 
to the one mentioned in the comments above.
We need the next idea: To get the components of a vector, say $v\in\mathbb R^n$, after in $\mathbb R^n$ had been set a basis change $B$, which is described by
$$b_1=a_{11}e_1+\cdots+a_{n1}e_n,$$
$$b_2=a_{12}e_1+\cdots+a_{n2}e_n,$$
$$...$$
$$b_n=a_{1n}e_1+\cdots+a_{nn}e_n,$$
then the associated matrix to that is
$$
[B]=\left[\begin{array}{cccc}
a_{11}&a_{12}&...&a_{1n}\\
a_{21}&a_{22}& & \\
\vdots&&\ddots\\
a_{n1}&a_{n2}&...&a_{nn}\\
\end{array}\right].$$
This matrix should be non singular to guaranty the linear independence of the $b_i$. Hence the new components for $v$ can be get by the multiplication $$B^{-1}v.$$
Once you know that let us consider a bi-linear map (as such, it represents 
a rank two tensor)
$$\mathbb R^n\times \mathbb R^n\stackrel Q\longrightarrow\mathbb R$$
defined by $v^{\top}Qw$. This is a generalization of a inner product. We are going to require that $Q$ be symmetric 
Now, one can ask for what is the effect on $Q$ when we change the basis,
and the situation lead to
\begin{eqnarray*}
v^{\top}Qw&=&(BB^{-1}v)^{\top}QBB^{-1}w,\\
&=&(B^{-1}v)^{\top}(B^{\top}QB)B^{-1}w.
\end{eqnarray*} 
From the last line of this equation you can see how the new components of $v$ and $w$ are paired with the matrix $B^{\top}QB$ and the value $v^{\top}Qw$ remains the same, that is invariant.
An important case for the choice of change of basis is the eigen-vectors of $Q$, because with them you are going to get the simpler version of the pairing $Q$ which gives for $B^{\top}QB$ a diagonal matrix. 
