Suppose that for each field $F$ a linear map $X(F): H\_M^{p,q}(F, \mathbb{Q}) \longrightarrow H\_M^{p,q}(F,\mathbb{Q})$ is given, such that $X$ commutes with inclusions of fields and transfers for finite field extensions. Is it proved, conjectured or doubted that each such map is just a multiplication on a rational number?



Do there exist nontrivial motivic cohomology operations preserving weights?
I agree with Mikhail Bondarko's comment, the condition in the question is certainly too weak. However, looking at cohomology operations (as the title suggests) something can be said. The basic issues involved in the question seem to be 


1. What can we say about stable operations for rational motivic cohomology?
2. How does the notion in the question (morphisms on cohomology compatible with restriction and transfers) relate to unstable or stable cohomology operations?


*For the first point:* if we consider stable operations in rational motivic cohomology, then a positive answer can be deduced from the work of Riou:


J. Riou. Algebraic K-theory, $\mathbb{A}^1$-homotopy and Riemann-Roch theorems. J. Top. 3 (2010), 229-264, [arXiv:0907.2710](http://arxiv.org/abs/0907.2710).


The description of endomorphisms of the K-theory spectrum in the stable homotopy category $\mathcal{SH}(S)\_{\mathbb{Q}}$ can be found in Section 5 of the arXiv version. The eigenspace decomposition in $\mathcal{SH}(S)\_{\mathbb{Q}}$ for Adams operations on K-theory gives rise to a spectrum representing motivic cohomology, cf. Definition 5.3.17. The stable operations of Beilinson motivic cohomology can be found in Remark 5.3.16, the component of bidegree $(0,0)$ is given by $K\_0(S)^{(0)}$. Over a base field, this means that all stable operations on rational motivic cohomology are given by scalar multiplication as required.


*About the second point,* I am not really sure. Topologically, I would use the identification $[K(A,n),K(B,n)]\cong \operatorname{Hom}(A,B)$ to see that a degree $0$ unstable cohomology operation is determined by its value on the point. 


An $\mathbb{A}^1$-homotopy version of this argument works with the notion of Eilenberg-Mac Lane spaces in Morel's book "$\mathbb{A}^1$-algebraic topology over a field". Assume $k$ is an infinite perfect base field. Then for each strictly $\mathbb{A}^1$-invariant sheaf of abelian groups, there is a corresponding Eilenberg-Mac Lane space, and an identification $[K(\mathcal{A},n),K(\mathcal{B},n)]\cong\operatorname{Hom}(\mathcal{A},\mathcal{B})$. On the right-hand side, we have morphisms of strictly $\mathbb{A}^1$-invariant sheaves of abelian groups. These sheaves are determined by their values over fields, and the same is true for morphisms. However, for a transformation $H^{p,q}(L,\mathbb{Q})\to H^{p,q}(L,\mathbb{Q})$ to extend to a morphism of the corresponding strictly $\mathbb{A}^1$-invariant sheaves of groups, it needs to be compatible with field inclusions and residue maps for valuations corresponding to codimension 1 points of smooth schemes. I think it follows from the theory developed in Morel's book that compatibility with transfers would follow from that. In particular, a slightly modified version of the notion of operation in the question gives rise to an actual map of Eilenberg-Mac Lane spaces (in the sense of Morel's book). 


There are two open points which I can't figure out right now. One is, if the spaces representing motivic cohomology are Eilenberg-Mac Lane spaces in the sense of Morel's book. If not, it would be more difficult to understand the relation between transformations over fields and cohomology operations. The other one is: even if that works, it is not clear to me how to express compatibility with suspension just by evaluation on fields.


