When does the identity $\|\mathrm{Tr}_B UXU^\dagger\|_1=\|\mathrm{Tr}_B X\|_1$, with $U$ unitary, hold? Let $\mathcal{H}_{AB}$ be a bipartite, complex Euclidean space, and let $U\colon\mathcal{H}_{AB}\to\mathcal{H}_{AB}$ be a unitary operator. Define the trace norm as
$$ \lVert X\rVert_1 = \text{Tr}(\sqrt{XX^\dagger}) $$
where $X$ is a linear operator on $\mathcal{H}_{AB}$, and $X^\dagger$ denote the complex conjugate of $X$. I know $\lVert UXU^\dagger\rVert_1 = \lVert X\rVert_1$, but does the following identity hold:
$$ \lVert \text{Tr}_B UXU^\dagger\rVert_1 = \lVert \text{Tr}_B X\rVert_1 $$
Assuming finite dimensionality is fine for my application. 
 A: If $X$ is positive, then $\|X\|_1=\operatorname{Tr}(X)$.
Moreover, if $X$ is positive, then also $\operatorname{Tr}_B(X)$ is. Therefore, if $X$ is positive, then $\|\mathrm{Tr}_B(X)\|_1=\operatorname{Tr}(X)$.
Finally, if $X$ is positive and $U$ unitary, also $UXU^\dagger$ is positive, and therefore
$$\|\mathrm{Tr}_B(UXU^\dagger)\|_1=\mathrm{Tr}(UXU^\dagger)=\operatorname{Tr}(X)=\|\mathrm{Tr}_B X\|_1.$$
The statement is false when $X$ is not positive. As a counter-example consider
$$X=\begin{pmatrix}
1 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}.$$
Then,
$$\operatorname{Tr}_B(X)=\begin{pmatrix}
2 & 0 \\ 0 & 0
\end{pmatrix}$$
and thus $\|\mathrm{Tr}_B(X)\|_1=2$.
However, taking
$$U = 
\begin{pmatrix}
 1 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 1 \\
\end{pmatrix},$$
we have
$$
UXU^\dagger =
\begin{pmatrix}
 1 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0
\end{pmatrix},
\qquad
\mathrm{Tr}_B(UXU^\dagger)=\begin{pmatrix}1&1\\0&1\end{pmatrix},
$$
and thus clearly
$$\|\mathrm{Tr}_B(UXU^\dagger)\|_1 \simeq 2.23
\neq 2=\|\mathrm{Tr}_B(X)\|_1.
$$
Following Norbert's suggestion in the comments, a better example using normal matrices might be given by a matrix of the form $X=A\otimes I/d$ with $A$ Hermitian and $U$ swapping the two spaces, so that $UXU^\dagger = I/d\otimes A$. Then, $\operatorname{Tr}_2(X)=A$,
$\|\mathrm{Tr}_2(X)\|_1=\|A\|_1$, 
whereas $\|\mathrm{Tr}_2(U XU^\dagger)\|_1=\operatorname{Tr}(A)$,
and $\operatorname{Tr}(A)\neq\|A\|_1$ unless $A\ge0$.
This question on math.SE might also be of interest.
