What is the difference between 3 qubits, 2 qutrits and a 6th level qunit? Are they equivalent? Why / why not?
Can 6 classical bits be super-densely coded into each?
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The Hilbert space dimension of $n$ qudits is $d^n$, where $d$ is the dimension of the qudit ($d=2$ for qubit, $d=3$ for qutrit, etc). So three qubits have an $8$ dimensional space, two qutrits have a $9$ dimensional space, and one $d=6$ qudit has a six dimensional space. As such, we cannot regard them as equivalent.
I guess you meant to compare situations with equal total Hilbert space dimension. Such as a comparing a pair of qubits with a $d=4$ system. In this case, there is mathematically no distinction. You could choose to relabel the basis states $|00\rangle$, $|01\rangle$, $|10\rangle$ and $|11\rangle$ as the qudit basis states $|0\rangle$, $|1\rangle$, $|2\rangle$ and $|3\rangle$. Then any qudit operations defined with the qudit basis could be equivalently defined with the qubits, and vice-versa. You could also use other mappings between basis states, this was just an example.
We could also use a subspace of a larger space to simulate a smaller one. For example, suppose you want to simulate a spin-$1$ particle, which is a 3 level system. You could do this using a pair of qubits (a four level system) and identifying three basis states of the former with three of the latter (such as $|-1\rangle$, $|0\rangle$ and $|1\rangle$ with $|00\rangle$, $|01\rangle$ and $|10\rangle$, for example). As long as you implement your spin-$1$ operations correctly, you'll always avoid the $|11\rangle$ state, and your two qubits effectively become a qutrit.
You might also be interested in my answer to the question What is the most economical and preferred basis for the qudit?
They are not equivalent. It can be seen by the fact that the system of $3$ qubits acts on a $8$ dimensional Hilbert space, the 2 qutrit system acts on a $9$ dimensional Hilbert space, and the 6 level qunit acts on a $6$ dimensional Hilbert space. Consequently, the nature of the states defined by each of the quantum systems is different.
This dimension argument comes from the fact that a k-level n-qunit system acts on a state space of dimension $k^n$.
For the superdense-coding, I am aware that Bell pairs are used in order to obtain the desired coding, and as you do not consider such entangled qubits, I am not sure to answer such question.