Is quantum cloning $|\psi\rangle|\psi_1\rangle|\psi_2\rangle|C\rangle\to e^{i\alpha}|\phi\rangle|\psi\rangle|\psi\rangle|C'\rangle$ prohibited? I think the no-cloning theorem is too restrictive, as in,
$$|\psi\rangle |\phi\rangle\to e^{i\alpha}|\psi\rangle|\psi\rangle \tag{1}$$
does not allow for any arbitrariness in the final state.
Instead, suppose we have three particles in initial states $|\psi \rangle$, $|\psi_1\rangle$ and $|\psi_2\rangle$,  and we have an external quantum system (called a cloner) in an initial state $|C\rangle$. The four are not mutually isolated. They can interact. We consider the cloning:
$$|\psi\rangle|\psi_1\rangle|\psi_2\rangle |C\rangle \to e^{i\alpha} |\phi \rangle |\psi\rangle|\psi\rangle |C'\rangle$$
We isolated the four components in the end. This gives us two copies of the state $|\psi\rangle$ and we get two arbitrary states $|\phi\rangle$ and $|C'\rangle$ that we don't care about.
The states of all four components of the system change in the process. In $(1) $, the state of the first particle remained unaffected, which is too much to ask for from an interaction process.
My question, is this type of cloning theoretically proven to be prohibited too?
 A: The essence of your question seems: Can we build a cloner if we allow for some extra quantum system in the output, i.e. a linear operation which does
$$
\lvert \psi\rangle\mapsto \lvert \psi\rangle\lvert \psi\rangle\lvert a_\psi\rangle\ ,
\tag{1}
$$
where $\lvert a_\psi\rangle$ is an ancilla which can depend on $\psi$, and which we trace out. (Such an operation could involve adding ancillas on the l.h.s.)
We can now follow precisely the normal argument, i.e. use linearity (using qubits):
$$
\begin{align}
\lvert 0\rangle & \mapsto \lvert 0\rangle\lvert 0\rangle\lvert a_0\rangle
\\
\lvert 1\rangle & \mapsto \lvert 1\rangle\lvert 1\rangle\lvert a_1\rangle\ .
\end{align}
\tag{2}
$$
Applying the cloner to $\lvert0\rangle+\lvert 1\rangle$ gives on the one hand (from (1))
$$
\lvert 0\rangle\lvert 0\rangle\lvert a_+\rangle
+\lvert 0\rangle\lvert 1\rangle\lvert a_+\rangle
+\lvert 1\rangle\lvert 0\rangle\lvert a_+\rangle
+\lvert 1\rangle\lvert 1\rangle\lvert a_+\rangle
$$
and on the other hand (from (2))
$$
\lvert 0\rangle\lvert 0\rangle\lvert a_0\rangle
+\lvert 1\rangle\lvert 1\rangle\lvert a_1\rangle\ .
$$
You will notice that the first expression has terms of the form
$\lvert 0\rangle\lvert 1\rangle\lvert \mathrm{stuff}\rangle$, while the second one doesn't. These terms cannot come from anything in the $\lvert\mathrm{stuff}\rangle$ register, since they are simply not contained in the span of $\lvert 0\rangle\lvert 0\rangle\lvert a_0\rangle$ and $\lvert 1\rangle\lvert 1\rangle\lvert a_1\rangle$.
Thus, the two expressions cannot be equal, regardless of what $\lvert a_\psi\rangle$ is (or what its dimension is), and thus, a linear cloning map of this form cannot exist.
Note that by Stinespring dilation, this also implies that there cannot be a completely positive $\mathcal E$ map which clones,
$$
\mathcal E(\lvert\psi\rangle\langle\psi\rvert) = 
\lvert\psi\rangle\langle\psi\rvert\otimes
\lvert\psi\rangle\langle\psi\rvert\ .
$$
