An intuitive explanation of the basis-independency of EPR pair As usual, we write the EPR pair as
$$
\frac{1}{\sqrt2}(\left|00\right> + \left|11\right>).
$$
A property of the EPR pair is that this definition is basis-independent, which means
$$
\frac{1}{\sqrt2}(\left|00\right> + \left|11\right>) = \frac{1}{\sqrt2}(\left|uu\right> + \left|vv\right>)
$$
as long as $\left|u\right>$ and $\left|v\right>$ are orthogonal in $\mathbb{R}^2$. (As pointed out by @ElioFabri, if we set $\left|u\right>=\left|0\right>$ and $\left|v\right>=i\left|1\right>$ then the equailty no longer holds. Hence we cannot put them in $\mathbb{C}^2$.)

Question: Is there any way to explain this other than an elementary direct computation?

 A: Here's a neat way to see this.
The main element is vectorization. Vectorization of a matrix $A=\sum A_{i,j}\vert i \rangle\langle j\vert$ is defined by $\vert\text{vec}({A})\rangle = \sum A_{i,j}\vert i \rangle\vert j\rangle$. Essentially, a bra turns into a ket. The cute thing about this is that you've now established a correspondence between matrices and bipartite pure states.
Next, consider what the identity matrix vectorizes to. You have $\sum_i \vert i \rangle\langle i\vert \rightarrow \sum_i \vert i \rangle\vert i\rangle $, which is an unnormalized Bell state. Indeed, if you chose to do it in two dimensions and the computational basis, you have your state $\vert 0\rangle\langle 0\vert + \vert 1\rangle\langle 1\vert \rightarrow \vert\Phi^+\rangle = \vert 00 \rangle + \vert 11\rangle$ upto normalization. You might be interested in the other Pauli matrices and what they do. I leave it to you to work it out but they simply give you the other Bell states. 
But wait, you're free to write the identity as $\sum_i\vert i \rangle\langle i\vert$ in any orthonormal basis. So indeed you can have $i = \{0,1\}$ or $i = \{+,-\}$ or as you wanted to write it, $i = \{u,v\}$ for some $\vert u\rangle \perp \vert v\rangle$. 
In some sense, the intuition is that because the identity matrix vectorizes to the $\vert\phi^+\rangle$ state, the freedom in representing the identity using any orthonormal basis carries over to the freedom in representing the vectorized state.
