In local class field theory, the reciprocity map is constructed by using the isomorphism ${\rm Br}(F)\simeq \mathbb{Q/Z}$, where $F$ is a local field and ${\rm Br}(F)$ is its Brauer group. The cohomological cup product $$ H^{0}({\rm Gal}(F^{\rm sep}/F),(F^{\rm sep})^{\times})\times H^{2}({\rm Gal}(F^{\rm sep}/F),\mathbb{Z})\longrightarrow {\rm Br}(F)$$ provides reciprocity map $F^{\times}\simeq H^{0}({\rm Gal}(F^{\rm sep}/F),(F^{\rm sep})^{\times})\longrightarrow {\rm Hom}(H^{2}({\rm Gal}(F^{\rm sep}/F),\mathbb{Z}),{\rm Br}(F))\simeq {\rm Gal}(F^{\rm ab}/F)$. For a finite field $\mathbb{F_{q}}$, we knew that $H^{2}({\rm Gal}(\mathbb{F_{q}^{\rm sep}/F_{q}}),\mathbb{Z})\simeq \mathbb{Q/Z}$. Recall that $\mathbb{Z}=K^{\rm top}_{0}(\mathbb{F}^{\rm sep}_{q})$ and $(F^{\rm sep})^{\times}\simeq K^{\rm top}_{1}(F^{\rm sep})$ where $F$ is a local field, we can predict that $H^{2}({\rm Gal}(F^{\rm sep}/F),K_{n}^{\rm top}(F^{\rm sep}))\simeq \mathbb{Q/Z}$ where $F$ is a $n$-dimensional local field. I have not shown this claim yet, but If this is true, I conjecture that we may be able to show higher class field theory by using cup product $$ H^{0}({\rm Gal}(F^{\rm sep}/F),K_{n}^{\rm top}(F^{\rm sep}))\times H^{2}({\rm Gal}(F^{\rm sep}/F),\mathbb{Z})\longrightarrow H^{2}({\rm Gal}(F^{\rm sep}/F),K_{n}^{\rm top}(F^{\rm sep})) \simeq \mathbb{Q/Z}. $$ So in this case, all I need is an isomorphism $H^{0}({\rm Gal}(F^{\rm sep}/F),K_{n}^{\rm top}(F^{\rm sep}))\simeq K_{n}^{\rm top}(F)$. This is clear when $n=1$. Ivan Fesenko proved that for a prime degree cyclic extension $L/F$ of a $n$-dimensional local field $F$, the invariant part $K_{n}^{\rm top}(L)^{{\rm Gal}(L/F)}$ is isomorphic to $K_{n}^{\rm top}(F)$.

Question. A finite Galois extension $E/F$ of $n$-dimensional local fields, is $0$-th cohomology $H^{0}({\rm Gal}(E/F),K_{n}^{\rm top}(E))$ isomorphic to $K_{n}^{\rm top}(F)$?