matrix elements of unitary operator I have read that the matrix elements of a unitary operator are the same in a pair of orthonormal basis such that {${|a_i\rangle}$} and {${|b_i\rangle}$} are orthonormal bases and ${|b_i\rangle} = U{|a_i\rangle}$:
$$\langle{a_i}|U|a_j\rangle = \langle{b_i}|U|b_j\rangle$$
I have 2 questions:
First, is the following computation right? We start with:
$$\langle{a_i}|U|a_j\rangle = \langle{a_i}|U1|a_j\rangle $$
and using $ {1} =UU^{\dagger} = U^{\dagger}U:$ 
$$\langle{a_i}|U|a_j\rangle = \langle{a_i}|UU^{\dagger}U|a_j\rangle $$
$$\langle{a_i}|U|a_j\rangle =\langle{a_i}U|U^{\dagger}|Ua_j\rangle $$
$$\langle{a_i}|U|a_j\rangle = \langle{b_i}|U^{\dagger}|{b_j}\rangle $$
Is this true? And if not, what did I do wrong?
Also, intuitively, why must it be that the matrix representation of a unitary operator looks the same in such bases?
Edit
The original question mistakenly started off: "I have read that the matrix elements of a unitary operator are the same in any pair of orthonormal basis" which as shown by celtschk in his answer is clearly false.
 A: The claim is false.
Counterexample: Be
$$U=\pmatrix{1&0\\0&-1}$$
This is easily checked to be an unitary matrix. In the orthonormal basis
$$\left|a_0\right>=\pmatrix{1\\0}, \left|a_1\right>=\pmatrix{0\\1}$$
we have $\left<a_i\middle|U\middle|a_j\right> = (-1)^i\delta_{ij}$.
The basis
$$\left|b_i\right> = \frac{1}{\sqrt{2}}\left(\left|a_0\right> + (-1)^i\left|a_1\right>\right)$$
can easily be checked to be an orthonormal basis as well, but
$$\begin{aligned}
\left<b_i\middle|U\middle|b_j\right>
&= \frac{1}{2}\left(\left<a_0\right| + (-1)^i\left<a_1\right>\right)U\left(\left|a_0\right> + (-1)^j\left|a_1\right>\right)\\
&= \frac{1}{2}\left(\left<a_0\middle|U\middle|a_0\right>
 + (-1)^i\left<a_1\middle|U\middle|a_0\right>
 + (-1)^j\left<a_0\middle|U\middle|a_1\right>
 + (-1)^{i+j}\left<a_1\middle|U\middle|a_1\right>\right)\\
&=\frac{1}{2}\left(1-(-1)^{i+j}\right)=1-\delta_{ij}
\end{aligned}$$
Clearly $(-1)^i\delta_{ij} \ne 1-\delta_{ij}$.
A simpler counterexample would be the above matrix with $\left|b_0\right> = \left|a_1\right>$ and $\left|b_1\right> = \left|a_0\right>$.
Indeed, the only matrices that look the same in every orthogonal bases are the multiples of the identity matrix.
