Same SVD for $U M V^{\dagger}$ and $M$ where $U$ and $V$ isometries: why? In the following document https://arxiv.org/pdf/1106.1445.pdf, on page 252 is stated the following property:
Let $U$ and $V$ be isometries. I emphasize the fact that they are not necesserally endomorphism, which mean: $U \in \mathcal{L}(H',H'')$ and $V \in \mathcal{L}(H,H_0)$ with $H_0 \neq H \neq H' \neq H''$ in all generality.
We have the following property for any operator $M \in \mathcal{L}(H,H')$ where $H$ and $H'$ are Hilbert spaces.
$$||U M V^{\dagger}||_1 = ||M||_1$$
The norm is defined via:
$$||M||_1=Tr(\sqrt{M^{\dagger} M})$$

My question:
The justification fo this property is just that $U M V^{\dagger}$ has the same singular value decomposition as $M$. It sounds like an obvious property but I don't get why it is true.
Indeed, in the case $U$ and $V$ are square matrices I would agree. But in a more general case where the input and output Hilbert space or not the same I don't get why it would be true.
I am ok with a simple proof in finite dimension. I am from quantum info/quantum physics side (I say this to avoid too mathematical abstract way to prove it if it exists)
 A: As I understood $U$ and $V$ are unitary operators on $H$ and $H'$ respectively (isometric linear operators on Hilbert spaces are exactly unitary operators, if we suppose that isometric implies bijective). In this case if $H$ and $H'$ are finite-dimensional it is simple to proceed by performing SVD (as you mentioned). I also want do describe another approach to unitary invariance (that is more clear for me and also works for trace-class operators on Hilbert spaces).
Consider approximative numbers $a_r(M) = d(M, L_r(H,H'))$ where $L_r$ is the set of all operators that have rank $\le r$ (distance is in usual operator norm). Theese are exactly the singular values of $M$ because of Eckart–Young–Mirsky theorem (https://en.wikipedia.org/wiki/Low-rank_approximation). So, approximative numbers give an alternative definition of singular values that is sometimes simplier to use. The sequence of approximative numbers is denoted by $a(M)$. Note that $M$ is compact iff $a(M) \in c_0$, $M$ is trace-class iff $a(M) \in l_1$, $M$ is Hilbert-Schmidt iff $a(M) \in l_2$ and $M$ has finite rank iff $a(M)$ vanishes for big enough indices (in finite-dimensional case all this situations coincide). Now also you can observe that $||M||_1 = ||a(M)||_1$ (i.e. norm of $a(M)$ in $l_1$) for all trace-class operators. And now the unitary invariance is obvious since multiplication by unitary operator preserves operator norm and preserves rank and therefore it preserves approximative (singular) numbers. That means that $a(UMV^\dagger) = a(M)$ for all unitary $U$ and $V$. And therefore $||UMV^\dagger||_1 = ||M||_1$.
Now let's proceed to "non-square" $U,V$. It appears that if $U$ and $V$ are only injective and isometric (norm-preserving) but not surjective then the proposition is also true. Let $V:H \rightarrow \hat{H}$ and $U:H' \rightarrow \hat{H'}$ be some isometric embeddings. Then there exist unitary isomorphisms $S: \hat{H} \rightarrow H \oplus \hat{H}_0$, $T: \hat{H'} \rightarrow H' \oplus \hat{H'_0}$ such that $SV = J_H$ and $TU = J_{H'}$ where $J_H:H \rightarrow H \oplus \hat{H}_0$ and $J_{H'}:H' \rightarrow H' \oplus \hat{H'}_0$ are canonical inclusions. Then $||UMV^\dagger||_1$ = $||TUMV^\dagger S^\dagger||_1 = ||J_{H'} M J_H^\dagger||_1$ since $T,S$ are unitary. But operator $\hat{M} = J_{H'}MJ_H^\dagger$ has very simple structure since $J_H^\dagger = PR_H:H \oplus \hat{H}_0 \rightarrow H$ - canonical projection. So we have an operator $\hat{M}:H \oplus \hat{H}_0 \rightarrow H' \oplus \hat{H'}_0$ that acts as $M$ on a direct summand $H$ and maps the direct summand $\hat{H}_0$ to zero. You can check (it is easy) that approximative numbers of $\hat{M}$ are equal to those of $M$. As for matrices you can observe that SVD of $M$ will give you the (I can be mistaken in terminology) truncated SVD of $\hat{M}$ where you omit zero singular values (you can think of $\hat{M}$ as a block $2\times 2$ matrix where left upper block is $M$ and other blocks are $0$).
A: I think I found a way. If you find a mistake please tell me.
The singular values of $U M V^{\dagger}$ are the squareroot of the positive eigenvalues of $(U M V^{\dagger})^{\dagger} (U M V^{\dagger})=V M^{\dagger} M V^{\dagger}$
It remains to prove that the eigenvalues of $V M^{\dagger} M V^{\dagger}$ are the same as the eigenvalues of $M^{\dagger} M$.
$V$ being an isometry, I don't have $V V^{\dagger} = I_{H_0}$, but I have $V^{\dagger} V=I_H$
Let $|\lambda\rangle$ be an eigenvector of $M^{\dagger} M$. I have:
$$ (V M^{\dagger} M V^{\dagger}) V |\lambda\rangle = \lambda V |\lambda \rangle$$
Thus, for any eigenvector $|\lambda\rangle$ of $M^{\dagger} M$, $V |\lambda \rangle$ is an eigenvector of $V M^{\dagger} M V^{\dagger}$ with the same eigenvalue.
It proves that $Sp(M^{\dagger} M) \subset Sp(V M^{\dagger} M V^{\dagger}) $

Reciprocally, let $|\epsilon\rangle$ be an eigenvector of $V M^{\dagger} M V^{\dagger}$.
I have:
$$V M^{\dagger} M V^{\dagger} |\epsilon\rangle = \epsilon |\epsilon\rangle$$
Thus, using $V^{\dagger} V = I_H$, I have:
$$M^{\dagger} M V^{\dagger} |\epsilon\rangle = \epsilon V^{\dagger}|\epsilon\rangle$$
Thus: $V^{\dagger} |\epsilon\rangle$ is an eigenvector of $M^{\dagger} M$ for the same eigenvalue.
Thus, I have $Sp(V M^{\dagger} M V^{\dagger}) \subset Sp(M^{\dagger} M)$

In the end I thus have: $Sp(V M^{\dagger} M V^{\dagger}) = Sp(M^{\dagger} M)$, which proves $||U M V^{\dagger}||_1=||M||_1$
