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; 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 $U^\dagger AU=B$.

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. 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 as well as this paper. 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!

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.

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 $T,S$ 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.

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    $\begingroup$ Has the conjecture been verified for small $n$? $\endgroup$ – Josiah Park Dec 20 '18 at 7:15
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    $\begingroup$ Good point @JosiahPark! I in fact have not been able to prove this even for $n=2$, but I edited my post to share some of my efforts in this simpler case. $\endgroup$ – Frederik vom Ende Dec 21 '18 at 8:40

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 $\mathrm{diag}(0,1,1)$.

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  • $\begingroup$ Splendid example, which (fun fact) even disproves an old result about vector-$d$-majorization when slightly modifying it: choose $d=(3,2,1)$ and $T(x,y,z):=(y+z,2x/3,x/3)$ so $Td=d$, $T$ is trace-preserving (i.e. preserves the sum of vector entries) but $T$ flips $(3,0,0)$ and $(0,2,1)$ which shows that $d$-majorization contrary to popular belief is not a partial ordering even if the entries of the weight vector $d$ are pairwise different. Anyways great work, thank you very much! $\endgroup$ – Frederik vom Ende Feb 7 '19 at 8:32

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