Isometric equivalence of purifications of quantum states Following the notes here (Quantum Information Theory Tips 5 at ETH), we state the following result. For any quantum state $\rho_A$ and purifications $\vert\psi\rangle_{AB}$ and $\vert\phi\rangle_{AC}$, there exists an isometry $V_{B\rightarrow C}$ such that $(I_A\otimes V_{B\rightarrow C})\vert\psi\rangle_{AB} = \vert\phi\rangle_{AC}$. Consider now $\rho_{A} = \frac{\mathbb{1}_A}{2}$, the maximally mixed state, and the following purifications.
$$|\psi\rangle_{A B}=\frac{1}{\sqrt{2}}\left(|0\rangle_{A}|+\rangle_{B}+|1\rangle_{A}|-\rangle_{B}\right) \quad \text{and} \quad|\phi\rangle_{A C}=\frac{1}{\sqrt{2}}\left(|0\rangle_{A}|000\rangle_{C}+|1\rangle_{A}|110\rangle_{C}\right)$$
Is it true that there is an isometry $V'_{C\rightarrow B}$ such that $(I_A\otimes V'_{C\rightarrow B})\vert\phi\rangle_{AC} = \vert\psi\rangle_{AB}$? Note that here $\text{dim}(\mathcal{H}_C) > \text{dim}(\mathcal{H}_B)$. If yes, how is this consistent with the following definition of isometries which state that they go from a smaller Hilbert space to a larger Hilbert space only? 

Let $\mathcal{H}$ and $\mathcal{H}^{\prime}$ be Hilbert spaces such
  that $\operatorname{dim}(\mathcal{H}) \leq$
$\operatorname{dim}\left(\mathcal{H}^{\prime}\right)$ An isometry $V$
  is a linear map from $\mathcal{H}$ to $\mathcal{H}^{\prime}$ such that
  $V^{\dagger} V=I_{\mathcal{H}}$. Equivalently, an isometry $V$ is a
  linear, norm-preserving operator, in the sense that
  $\||\psi\rangle\left\|_{2}=\right\| V|\psi\rangle \|_{2}$ for all
  $|\psi\rangle \in \mathcal{H}$.

This is related to my previous question here but I am still not sure about this dimensional problem.
 A: The essence is the following: You can write any purification in Schmidt form (note that this is not a transformation, just rewriting the state in a different basis).  Then, any two purifications of a given state will be of the form
$$
|\psi\rangle = \sum \lambda_i |a_i\rangle \otimes |b_i\rangle \in \mathcal H_A\otimes \mathcal H_B$$
and
$$
|\phi\rangle = \sum \lambda_i |a_i\rangle \otimes |c_i\rangle \in \mathcal H_A\otimes \mathcal H_C\ .
$$
To relate the two purifications, you must construct a transformation which maps the orthogonal set of vectors $\{|b_i\rangle\}$ to the orthogonal set of vectors $\{|c_i\rangle\}$. 
Restricted to the span of those vectors, this is a (unique!) unitary transformation.  If either $\mathcal H_A$ or $\mathcal H_B$ is bigger than the span, you can pad this transformation such that it still has orthogonal rows or columns (depending which dimension is bigger), such that one of them is an isometry -- the one from the smaller to the bigger space -- and the converse transformation correspondingly a partial isometry, or the dagger of an isometry.
A: As far as I can tell the link does not state that

For any quantum state $\rho_A$ and purifications $\vert\psi\rangle_{AB}$ and $\vert\phi\rangle_{AC}$, there exists an isometry $V_{B\rightarrow C}$ such that $(I_A\otimes V_{B\rightarrow C})\vert\psi\rangle_{AB} = \vert\phi\rangle_{AC}$.

Which is just as well since that claim is incorrect, as your example proves!
It states

any two purifications are equivalent up to an isometry on the purifying system

which is a much more reasonable claim. In particular for the two states to be "equivalent up to an isometry on the purifying system" all that is required is that either there exists an isometry $V:B\to C$ or an isometry $V:C\to B$. It is not necessary that there are isometries both ways. As I mentioned in my comment there is an isometry both ways if and only if the two isometries are unitaries and both the spaces have the same dimension.
Note that the adjoint of an isometry is not an isometry, let alone an inverse of the isometry you started with. In general an isometry consists of an extension of your Hilbert space (i.e. adding some extra dimensions) followed by doing a unitary. The natural "inverse" operation to adding extra dimensions is the partial trace, but this is certainly not isometric (or useful here).
A: An isometry is a map such that
$$ \langle Vx,Vy\rangle=\langle x,y\rangle$$
if the image of $V$ has smaller dimension than its domain, then clearly this property cannot hold, as if we have an orthonormal basis 
$$ \langle x_i,x_j\rangle=\delta_{ij}$$
we cannot have 
$$\langle Vx_i,Vx_j\rangle=\delta_{ij}\tag{$*$} $$
because there aren't enough orthogonal vectors in the image of $V$. Instead you can have a partial isometry, i.e. a map $V$ such that $(*)$ holds for a subset $\{x_j\}_{j=1}^{d_V}$ where $d_V$ is the dimension of the image of $V$, and that sends the other vectors to $0$. In practice this means projecting your initial space onto a subspace of the same dimension as the image of $V$ and then applying an isometry.
More precisely, a partial isometry is a map that is an isometry on the orthogonal complement of its kernel.
what ort1426 says is correct and enough in my opinion, this already shows isometric equivalence, but a more complete statement could be

Let $|\psi\rangle_{AB}$ and $|\psi'\rangle_{AC}$ be two purifications of $\rho_A$. Then there exists a partial isometry $V_{B\to C}$ such that $V|\psi\rangle=|\psi'\rangle$

You already know how to prove case where $\mathrm{dim}(B)\leq \mathrm{dim}(C)$, then $V$ is an isometry or a unitary (which are a special case of partial isometry, despite the names), if $\mathrm{dim}(B)> \mathrm{dim}(C)$, consider a Schmidt decomposition of $|\psi\rangle$ and $|\psi'\rangle$
$$ |\psi\rangle_{AB}=\sum_{k=1}^{r} s_k |\alpha_k\rangle|\beta_k\rangle\\|\psi'\rangle_{AC}=\sum_{k=1}^{r} s_k |\alpha_k\rangle|\beta_k'\rangle$$
the $\alpha_k$ are equal because the states must both partial trace to $\rho_A$. We clearly have $r<\mathrm{dim}(C)$. Extend the $|\beta_k\rangle$ to a basis of $B$ arbitrarily and define
$$ V_{B\to C}|\beta_k\rangle=\begin{cases} |\beta_k'\rangle \quad &\textrm{if } k\leq r\\ 0 \quad &\textrm{otherwise}
\end{cases}$$
$V$ is a partial isometry and has the desired property, basically, you didn't need such a big Hilbert space to begin with, as the rank of the Schmidt decomposition is smaller than the dimension of your auxiliary space anyway, and $V$ throws away by projection the useless dimensions.
