Let $F$ be an padic field, and $E$ be a quadratic extension of $F$, then is $({F^{\times})^{ diag}}\backslash(GL_2 \times E^{\times})_{det=\mathbb{N}}$ isomorphic to some unitary group $U_{E/F}(2)$? Here, $\mathbb{N}(e)=e *\sigma(e)$, where $\sigma$ is the nontrivial element in $Gal(E/F)$. And $(GL_2 \times E^{\times})_{det=\mathbb{N}} :=\{(g,e)\in GL_2 \times E^{\times}det(g)=\mathbb{N}(e)\}$.

3$\begingroup$ Can you explain the notation ${\rm det}={\mathbb N}$? $\endgroup$ – Paul Broussous Jan 8 at 18:13

$\begingroup$ Does "det=\mathbb N" mean that the determinant is a norm, by any chance? $\endgroup$ – paul garrett Jan 8 at 22:33

$\begingroup$ Also, I think $U_{E/F}(2)$ is not reliably universal notation. I'm guessing that you mean the hermitian form to be two diagonal 1's, and conjugation is Galois conjugation? $\endgroup$ – paul garrett Jan 8 at 22:34

$\begingroup$ I have edited the problem again. $\endgroup$ – Cooler Panda Jan 9 at 16:09

$\begingroup$ The answer depends on whether you regard $({F^{\times})^{ diag}}\backslash(GL_2 \times E^{\times})_{det=\mathbb{N}}$ as an algebraic group over $F$ or as a topological group (the group of points). $\endgroup$ – Mikhail Borovoi Jan 9 at 20:58
I cannot answer this question for now. However, I have made some relevant calculations that are too long for a comment.
We set $$G_0=(GL_2 \times E^{\times})_{{\rm det}={\rm Nm}}\,,\quad G_1=G_0/(F^\times)^{\rm diag},\quad G_2=U_{E/F}(1,\lambda),$$ where $\lambda\in F^\times$ and $U_{E/F}(1,\lambda)$ denotes the unitary group of the diagonal hermitian form $$h(z_1,z_2)={\rm Nm}(z_1)\lambda{\rm Nm}(z_2).$$
We wish to know whether $G_1$ and $G_2$ are isomorphic. If $\lambda\notin{\rm im}({\rm Nm})$, then the hermitian form $h$ does not represent 0, the topological group $G_2$ is compact, and hence, it is not isomorphic to the noncompact group $G_1$ (this has been already noticed by Paul Broussous). Therefore, let us assume that $\lambda\in{\rm im}({\rm Nm})$; then we may assume that $\lambda=1$ and $G_2=U_{E/F}(1,1)$.
Consider the homomorphism $$\alpha\colon G_0\to F^\times,\quad (g,e)\mapsto {\rm Nm}(e).$$ Clearly, ${\rm im}(\alpha)={\rm im}({\rm Nm})\subset F^\times$. The homomorphism $\alpha$ induces a homomorphism $$\beta\colon G_1\to F^\times/F^{\times\,2},$$ and we have $${\rm im}(\beta)= {\rm im}({\rm Nm})/F^{\times\,2}.$$ We set $$U_1=\{e\in E\ \ {\rm Nm}(e)=1\}.$$
We have a canonical homomorphism $$\gamma\colon{\rm SL}(2,F)\times U_1\to G_1\quad (g,e)\mapsto [g,e],$$ where $[g,e]$ denotes the class of the pair $(g,e)\in G_0$. Then $$\ker(\gamma)=\{\pm 1\},\quad {\rm im}(\gamma)=\ker(\beta).$$ Thus we obtain a short exact sequence $$1\to ({\rm SL}(2,F)\times U_1)/\{\pm 1\}\to G_1\to {\rm im}({\rm Nm})/F^{\times\,2}\to 1.$$
On the other hand, we have a canonical homomorphism $$\delta\colon {{\rm SU}_{E/F}}(1,1)\times U_1\to G_2 \quad (s,e)\mapsto se,$$ whose kernel is $\{\pm 1\}$ and whose cokernel is $U_1/U_1^2$, where we write $U_1^2$ for the subgroup of squares in $U_1$. Thus we obtain a short exact sequence $$ 1\to({{\rm SU}_{E/F}}(1,1)\times U_1)/\{\pm 1\}\to G_2\to U_1/U_1^2\to 1.$$ We have ${{\rm SU}_{E/F}}(1,1)\simeq{\rm SL}(2,F)$ (see, for instance, the book on classical groups by Jean Dieudonné or the book Geometric Algebra by Emil Artin). Thus we obtain a short exact sequence $$ 1\to({\rm SL}(2,F)\times U_1)/\{\pm 1\}\to G_2\to U_1/U_1^2\to 1.$$
Comparing the short exact sequences for $G_1$ and for $G_2$, we see that if the 2groups ${\rm im}({\rm Nm})/F^{\times\,2}$ and $U_1/U_1^2$ are nonisomorphic (that is, have different orders), then $G_1$ and $G_2$ are not isomorphic "in a nice way".
I do not know whether the 2groups ${\rm im}({\rm Nm})/F^{\times\,2}$ and $U_1/U_1^2$ are isomorphic or not. I suggest for OP to ask this on MathOverflow.