Does the F-unitary group isomorphism arises from a conformal isometry?

Question

Let $K$ be a CM-field with totally real subfield $F$. Let $(V_1, h_1)$ and $(V_2, h_2)$ be two $n$-dimensional $K$-vector spaces with nondegenerate Hermitian forms, where $n\geq 3$.

Question 1 Does every $F$-group isomorphism $PU(V_1, h_1)\cong PU(V_2, h_2)$ arise from an isometry of $(V_1, h_1)\cong (V_2, \lambda h_2)$ for some $\lambda\in F^{\times}$ ?

Edit. Question 2 Suppose there exists an $F$-group isomorphism $PU(V_1, h_1)\cong PU(V_2, h_2)$, can we deduce $(V_1, h_1)\cong (V_2, \lambda h_2)$ for some $\lambda\in F^{\times}$, or any other relations between $h_1$ and $h_2$?

A related statement for orthogonal group is given in the comment to the question $p$-adic orthogonal groups in four variables .

We are interested in the case that $K$ is a cyclotomic field and $h$ has signature $(1,n-1)$ for an embedding $F\to \mathbb{R}$ and is definite for all the other embeddings.

Answer 1

The answer to Question 1 is No. Indeed, take $V_1=V_2=V:=K^n$ and write $$ h_1(z)=h_2(z)=h(z):=\lambda_1 z_1 \bar z_1+\dots+\lambda_n z_n\bar z_n\quad\text{with}\ \lambda_i\in F^\times.$$ Write $\widetilde G=U(V,h),\ G={\rm PU}(V,h)$. Define $$\tilde\sigma\colon\,\widetilde G\to \widetilde G,\quad g\mapsto \bar g\quad\text{for}\ g\in \widetilde G(F)\subset {\rm GL}(n,K).$$

Then the $F$-automorphism $\tilde \sigma$ of $\widetilde G$ induces an $F$-automorphism $\sigma$ of $G$. Since $n\ge 3$, this autmorphism $\sigma$ is outer, and hence it is not a conformal isometry. The difference with the case of Brian Conrad's comment is that in his case the algebraic group has no outer automorphisms.

Edit. The answer to Question 2 is Yes.

The isomorphism classes of twisted forms $(V',\,F^\times\cdot h')$ of the Hermitian space $(V,\,F^\times\cdot h)$ bijectively correspond to $H^1(F, {\rm GU}(V,h))$, where ${\rm GU}(V,h)={\rm Aut}(V,\,F^\times\cdot h)$. The isomorphism classes of twisted forms of the algebraic $F$-group ${\rm PU}(V,h)$ bijectively correspond to $H^1(F,{\rm Aut}({\rm PU}(V,h)))$. We wish to show that the kernel $$\ker\big[ H^1(F, {\rm GU}(V,h))\longrightarrow H^1(F,{\rm Aut}({\rm PU}(V,h)))\big]$$ is trivial.

We factor the above arrow as $$H^1(F, {\rm GU}(V,h))\to H^1(F,{\rm Aut}({\rm PU}(V,h))^0)\to H^1(F,{\rm Aut}({\rm PU}(V,h)))$$ (where $^0$ denotes the identity component), and show that both arrows have trivial kernels. For the first arrow we have an exact sequence $$1=H^1(F,K^\times)\to H^1(F, {\rm GU}(V,h))\to H^1(F,{\rm Aut}({\rm PU}(V,h))^0),$$ which shows that the kernel is trivial. For the second arrow we have a short exact sequence of $F$-groups, $$1\to {\rm Aut}({\rm PU}(V,h)^0)\to {\rm Aut}({\rm PU}(V,h))\to {\rm Aut}(A_{n-1})\to 1,$$ where $A_{n-1}$ is the corresponding Dynkin diagram. We obtain a cohomology exact sequence \begin{align*} {\rm Aut}({\rm PU}(V,h))(F)\to {\rm Aut}(A_{n-1})(F)\to H^1(F,&{\rm Aut}({\rm PU}(V,h))^0)\\ &\to H^1(F,{\rm Aut}({\rm PU}(V,h))) \end{align*} where the arrow ${\rm Aut}({\rm PU}(V,h))(F)\to {\rm Aut}(A_{n-1})(F)$ is surjective by the answer to Question 1. This shows that the arrow $$H^1(F,{\rm Aut}({\rm PU}(V,h))^0)\to H^1(F,{\rm Aut}({\rm PU}(V,h)))$$ has trivial kernel, which completes the proof.