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When we say that a tensor is an array of numbers that transform according to some formula from one basis to another, can both bases be of the same coordinate system?

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Yes, for sure.

Indeed for 1 rank tensor we have that $A^{i} = \frac{\partial x^i}{\partial x^{\prime j} } A^{\prime j} $.

Even looking at this formula we can say that there is no such constrains on coordinates system, as you assumed. Problem occur when you have non-differentiable functions (derivative does not exist ...)

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When we say that a tensor is an array of numbers that transform according to some formula from one basis to another, can both bases be of the same coordinate system?

Nobody says that. You are making confusion between a tensor and its representations onto some bases.

Given $V$ as a vector space and $V^* $ as its dual, a tensor of type $(r,s)$ is, by definition, any multilinear map $$ \tau\colon V^r\times{V^*}^s\to \mathbb{F} $$ $\mathbb{F}$ being any field. Chosen a basis $\left\{\textbf{e}_i\right\} \in V$ and its corresponding dual $\left\{\alpha^j\right\} \in V^*$ such that $\alpha^j(\textbf{e}_i)=\delta^j_i$ the components of the tensor with respect to these bases are the values of the multilinear map $\tau$ evaluated thereupon. Standard matrix multiplication rules for components follows by distributivity of the product and the rules for the change of basis.

As an example for $(r,s)=(1,1)$ we have $$ \tau\colon V\times V^*\to\mathbb{R} $$ with $\tau_i^j=\tau(\textbf{e}_i,\alpha^j)$. Given a change of basis as orthogonal matrix $O$ we have $$ \tau'(\textbf{e}_i',\alpha'^j)=\tau(O_i^k\textbf{e}_k,\,{O^{-1}}^j_r\alpha^r)= O_i^k\,{O^{-1}}^j_r\,\tau_k^r $$ which terminates the proof. As you see, should you choose the same basis, the transformation matrix will reduce to the identity.

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    $\begingroup$ "Nobody says that." Sadly, you would be surprised. "A tensor is a thing that transforms like a tensor" is disturbingly common. $\endgroup$ – JohnnyMo1 Jul 2 '15 at 4:05

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