The following problem which has been on my mind for a while now arises from the realm of quantum information involving quantum channels with a common fixed point of full rank, as well as majorization theory, but can really be boiled down to a problem in linear algebra.


One considers maps acting on complex $n\times n$ square matrices which are linear, trace-preserving and positive *(more precisely "positivity-preserving", i.e. if $A\in\mathbb C^{n\times n}$ is positive semi-definite, then so is its image under the map in question)*. Let us denote the convex set of all such maps by $P(n)\subset\mathcal L(\mathbb C^{n\times n})$ where the latter is the vector space of all linear maps/operators on $\mathbb C^{n\times n}$.



> 
> **Conjecture.** Let $A,B,X\in\mathbb C^{n\times n}$ with $A,B$ hermitian and $X$ positive definite. If there exist $T\_1,T\_2\in P(n)$ with $$T\_1(X)=T\_2(X)=X,\quad T\_1(A)=B\quad\text{and}\quad T\_2(B)=A,$$ then there exists unitary $U\in\mathbb C^{n\times n}$ such that $[U,X]=0$ and $U^\dagger AU=B$. 
> 
> 
> 


The assumption of a common fixed point is essential. To see this, consider arbitrary $\rho\_1,\rho\_2\in\mathbb C^{n\times n}$ positive semi-definite and of trace $1$. The maps $T\_i:\mathbb C^{n\times n}\to \mathbb C^{n\times n}$, $X\mapsto \rho\_{2-i} \operatorname{tr}(X)$ are in $P(n)$ and satisfy $T\_1(\rho\_1)=\rho\_2$, $T\_2(\rho\_2)=\rho\_1$, but in general $\rho\_1,\rho\_2$ (of course) are not unitarily equivalent. So the common fixed point seems to rule out such "projective" maps from $P(n)$ which only have one-dimensional range. 


Note that maps of the above form for example come up in [majorization theory](https://www.sciencedirect.com/science/article/pii/0024379589905806); for two hermitian matrices $A,B$ one can show that $A$ majorizes $B$ (comparison of the vectors of eigenvalues) if and only if there exists $T\in P(n)$ with $T(\operatorname{id})=\operatorname{id}$ and $T(A)=B$, cf. e.g. Theorem 7.1 in the linked manuscript. In such a framework *(that is, $X=\operatorname{id}$ and $A,B$ hermitian)*, one quickly sees that the above conjecture in fact holds as then $\sigma(A)=\sigma(B)$. The arising difficulty for $X\neq \operatorname{id}$ in some sense comes from the loss os symmetry: a map $T\in\mathcal L(\mathbb C^{n\times n})$ is trace-preserving if and only if its dual $T^\*$ (w.r.t. the Hilbert-Schmidt inner product) is unital/identity-preserving. Thus one may reformulate the conjecture as follows.



> 
> **Conjecture (v2).** Let $A,B,X\in\mathbb C^{n\times n}$ with $A,B$ hermitian and $X$ positive definite. If there exist positive, linear maps $T\_1,T\_2:\mathbb C^{n\times n}\to \mathbb C^{n\times n}$ such that $$T\_1(X)=T\_2(X)=X,\quad T\_1^\*(\operatorname{id})=T\_2^\*(\operatorname{id})=\operatorname{id},\quad T\_1(A)=B\quad\text{and}\quad T\_2(B)=A,$$ then there exists unitary $U\in\mathbb C^{n\times n}$ such that $[U,X]=0$ and $U^\dagger AU=B$.
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> 
> 


One readily verifies that there is a one-to-one connection between unital maps from $P(n)$ (as used in majorization theory) and positive, linear maps satisfying $T(Y)=Y$ and $T^\*(Y^{-1})=Y^{-1}$ for some $Y>0$ via the transformation
$$
\Psi:\mathcal L(\mathbb C^{n\times n})\to P(n)\qquad T\mapsto Y^{-1/2}T(Y^{1/2}(\cdot)Y^{1/2})Y^{-1/2}
$$
However, I do not see if and how this could be connected to the kind of maps from Conjecture (v2):
$$
T(\operatorname{id})=T^\*(\operatorname{id})=\operatorname{id} \overset\checkmark\leftrightarrow T(X^{1/2})=X^{1/2},T^\*(X^{-1/2})=X^{-1/2} \overset?\leftrightarrow T(X)=X,T^\*(\operatorname{id})=\operatorname{id}
$$


As a final remark to this post: the above conjecture, if true, seems to be closely related to Wigner's theorem for state-automorphisms${}^1$ or more generally [linear isometries between matrix algebras](https://mathoverflow.net/a/310497/116991). I am also aware of results on quantum channels and full-rank fixed points, cf. Chapter 6.4 in [this script on quantum channels & operations by M. Wolf](https://www-m5.ma.tum.de/foswiki/pub/M5/Allgemeines/MichaelWolf/QChannelLecture.pdf) as well as [this paper](https://aip.scitation.org/doi/10.1063/1.5027616). However, so far I was not yet able to solve this puzzle. Thanks in advance for any answer or comment!


*P.S.* I hope the tags I chose for this question are fitting - but I am of course welcoming any comment addressing whether the tags are suitable or not, or if they can be extended in a meaningful way!




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Footnote ${}^1$: Another idea could be, starting from $T\_1,T\_2$, construct **one** map in $P(n)$ which maps $A$ to $B$, $B$ to $A$, $X$ to $X$ and acts bijectively on $\operatorname{span}\lbrace A,B,X\rbrace$ or perhaps on the convex, compact set of all density matrices (positive semi-def. & trace $1$). However, I do not see yet how one would go on about such a construction.




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*Edit:* Due to the comment of Josiah Park, let me share my efforts on the case $n=2$. Let $A,B,X\in\mathbb C^{2\times 2}$ such that $A,B$ hermitian and $X>0$. Note that $T(A)=B$ (as well as $S(B)=A$) trivially implies $\operatorname{tr}(A)=\operatorname{tr}(B)=:c\in\mathbb R$ so using linearity of $T,S$ and the common fixed point $X$, we may go to the trace-less hyperplane of hermitian matrices via
$$
T\big(\underbrace{A-\tfrac{c}{\operatorname{tr}(X)}X}\_{=:\hat A}\big)=\underbrace{B-\tfrac{c}{\operatorname{tr}(X)}X}\_{=:\hat B}\qquad S\big(B-\tfrac{c}{\operatorname{tr}(X)}X\big)=A-\tfrac{c}{\operatorname{tr}(X)}X\,.
$$
As $\hat A,\hat B$ are hermitian and trace-less, there exist $a,b\geq 0$ and unitaries $V\_a,V\_b$ such that $\hat A=V\_a\operatorname{diag}(a,-a)V\_a^\dagger$ and $\hat B=V\_b\operatorname{diag}(b,-b)V\_b^\dagger$. Due to Theorem 3.1 in [this paper by Perez-Garcia, Wolf, Petz and Ruskai](https://aip.scitation.org/doi/10.1063/1.2218675) $T,S$ for $n=2$ are $p$-norm contractive on the trace-less hermitian hyperplane for all $p\in[1,\infty]$. This yields
$$
\|\hat B\|\_\infty= \|T(\hat A)\|\_\infty\leq \|T\|\_{\infty\to\infty}\|\hat A\|\_\infty =\|\hat A\|\_\infty = \|S(\hat B)\|\_\infty\leq \|S\|\_{\infty\to\infty}\|\hat B\|\_\infty =\|\hat B\|\_\infty
$$
so $a=\|\hat A\|\_\infty=\|\hat B\|\_\infty=b$ and $\hat A,\hat B$ are unitarily equivalent. This translates to the existence of unitary $U$ with
$$
U^\dagger AU-\tfrac{c}{\operatorname{tr}(X)}U^\dagger XU=U^\dagger \big(A-\tfrac{c}{\operatorname{tr}(X)}X\big) U=U^\dagger \hat A U=\hat B=B-\tfrac{c}{\operatorname{tr}(X)}X
$$
so
$$
U^\dagger AU=B-\tfrac{c}{\operatorname{tr}(X)}U^\dagger [U,X]\,.
$$
If $X\propto\operatorname{id}$ we are done. However, if $X\not\propto\operatorname{id}$ then one yet has to show that the non-empty set
$$
\lbrace U\in\mathcal SU(2)\,|\, U^\dagger AU=B-\tfrac{c}{\operatorname{tr}(X)}U^\dagger [U,X]\,\rbrace
$$
also is non-empty under the additional constraint $[U,X]=0$ - or maybe the original $U$ for some reason already satisfies $[U,X]=0$? This I sadly do not see at the moment - on the other hand, the full-rank condition of $X$ did not yet come into play it seems as we only used $X$ positive semi-definite & $X\neq0$.


Anyways, this approach is very explicit as in low dimensions such calculations are quite simple and contractivity of positive, trace-preserving maps in the $\infty$-/operator norm holds for $n=2,3$ (but fails for $n>3$; however, all counter-examples to this I know of do not have any matrix of full rank in their range). Also given how I used it, I am not sure how useful the operator norm for larger $n$ is regardless.



Positive and trace-preserving transformations with a common fixed point of full rank
No, even on $D\_3:={\mathbb C} \oplus {\mathbb C} \oplus {\mathbb C}$ (the diagonal of $M\_3({\mathbb C})$. Let $e\_0,e\_1,e\_2$ denote the standard basis for $D\_3$ and put $T\colon e\_0\mapsto \frac{1}{2}(e\_1 + e\_2),\, e\_1\mapsto e\_0,\,e\_2\mapsto e\_0$ on $D\_3$ (or on $M\_3({\mathbb C})$ by ignoring the off-diagonal entries). Then, $T$ is a trace-preserving positive map which fixes $\mathrm{diag}(2,1,1)$ and flips $\mathrm{diag}(2,0,0)$ and $\mathrm{diag}(0,1,1)$.


