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