How to create a quantum algorithm that produces 2 n-bit sequences with equal number of 1-bits? I am interested in a quantum algorithm that has the following characteristics:

*

*output = 2n bits  OR 2 sets of n bits (e.g. 2 x 3 bits)

*the number of 1-bits in the first set of n-bits must be equal to the number of 1-bits in the second set.  E.g. correct output =  0,0,0, 0,0,0   (both 3-bit sets have zero 1-bits); 1,0,0, 0,1,0   (both 3-bit sets have one 1-bit); 1,1,0, 0,1,1   (both 3-bit sets have two 1-bit)

*Each time the quantum algorithm runs it must randomly return one of the possible solutions.  There are 2 good ways to interpret "randomly return one of the possible solutions":  (1) each possible good solution has equal chance of being returned by the quantum algorithm. (2) every possible good solution has a chance > 0 of being returned.

Any idea how I can best implement such an algorithm on a quantum computer ?
FYI I have tried the following algorithm (where n = 2 ) but it missed the 2 answers 0110 and 1001.
screenshot of the quantum circuit + simulator output
 A: Here is one way to achieve this, for concreteness described for $n=2$: Start with two registers of $2$ qubits, initialised as $|00\rangle|00\rangle$; apply a Hadamard transformation to each of the qubits in the first register, resulting in
$$(|00\rangle+|10\rangle+|01\rangle+|11\rangle)|00\rangle$$
(I leave out the normalisation factor). Then apply a sequence of CNOT gates with qubit $p\in\{1,2,3\}$ from register 1 as control and qubit $p$ from register 2 as target. This produces the entangled state
$$|00\rangle|00\rangle+|11\rangle|11\rangle+|10\rangle|10\rangle+|01\rangle|01\rangle$$
To obtain also the permutations $|10\rangle|01\rangle$ and $|01\rangle|10\rangle$ perform a square-root-of-SWAP operation $\sqrt{\text{SWAP}}$ on the qubits in the second register.
recall that $\sqrt{\text{SWAP}}|\alpha\beta\rangle=\frac{1}{\sqrt{2}}\left(|\alpha\beta\rangle+i|\beta\alpha\rangle\right)$
The resulting state is
$$(1+i)|00\rangle|00\rangle+(1+i)|11\rangle|11\rangle+
|10\rangle(|10\rangle+i|01\rangle)+|01\rangle(|01\rangle+i|10\rangle)$$
Finally measure the qubits in both registers to obtain a random output. The weights are such that there is equal probability to obtain one of the even parity terms (0000, 1111) as one of the odd parity terms (1010, 1001, 0101, 0110).
