Let $X_N$ denote the Fermat curve defined over $\mathbb{Q}$ by the equation $x^N+y^N-z^N=0$ and let $X_{N,\mathbb{Q}(\mu_N)}$ be the base change. Let $G$ be the Galois group of $\mathbb{Q}(\mu_N)/\mathbb{Q}$, hence $G \cong (\mathbb{Z}/N\mathbb{Z})^\times$. Consider the motivic cohomology $H^2_\mathcal{M}(X_N,\mathbb{Q}(2))$. I would like to know whether

$$ H^2_\mathcal{M}(X_N,\mathbb{Q}(2)) \cong H^2_\mathcal{M}(X_{N,\mathbb{Q}(\mu_N)},\mathbb{Q}(2))^G.$$

**Motivation:**

Let $W:=(\mathbb{Z}/N\mathbb{Z})^2$ act on $X_{N,\mathbb{Q}(\mu_N)}$ as

$$(r,s)(x:y:z) := (\zeta^rx:\zeta^sy:z),$$

where $\zeta$ is a primitive $N-$th root of unity. I am reading On the regulator of Fermat motives and generalized hypergeometric functions by N. Otsubo, where he proves some results of surjectivity of regulators using the splitting of the motive $h^1(X_N)$ into motives $X_N^{[a,b]}=(X_N,p_N^{[a,b]})$, where $p_N^{[a,b]}$ is in $\mathbb{Q}[W]^G$. I am working on something related, but I am not much acquainted with motivic cohomology. My definitions (which are quite *ad hoc* for this situation) come from Milnor $K-$theory:

$$H^2_\mathcal{M}(X_N,\mathbb{Q}(2)) \cong K_2(X_N)_\mathbb{Q} := \ker\left( K_2^M(k(X_N))\otimes \mathbb{Q} \overset{T\otimes\mathbb{Q}}{\longrightarrow} \bigoplus_{x \in X_N(\overline{\mathbb{Q}})} \overline{\mathbb{Q}}^\times \otimes \mathbb{Q} \right),$$ where $T$ denotes the Tame symbol (see for instance page 27 of the above article). My original problem is that I want to show that if $e_N$ is in $K_2(X_N)_\mathbb{Q}$, then the elements $e_N^{[a,b]}:=p_N^{[a,b]}(e_N)$ defined by Otsubo are in $K_2(X_N)_\mathbb{Q}$ as well. But, in principle, I only know that they are in $K_2(X_{N,\mathbb{Q}(\mu_N)})_\mathbb{Q}^G$. So I would like to know whether the two vector spaces are equal. The problem is that, starting from my definition, I think that we have

$$k(X_N)^\times \otimes k(X_N)^\times \subsetneqq (k(X_{N,\mathbb{Q}(\mu_N)})^\times \otimes k(X_{N,\mathbb{Q}(\mu_N)})^\times)^G.$$

So if things work out in the end it must be thanks to the Steinberg relation $a\otimes(1-a)$ that still has to be quotiented out in order to define $K_2(k(X_N))$, or thanks to the Tame symbol. But this seems to be difficult to check, hence I was wondering if there is some general argument for the equality

$$K_2(X_N)_\mathbb{Q}=K_2(X_{N,\mathbb{Q}(\mu_N)})_\mathbb{Q}^G$$

for example coming from motivic cohomology.