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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 $$1=H^1(F,K^\times)\to H^1(F, {\rm GU}(V,h))\to H^1(F,{\rm Aut}({\rm PU}(V,h))),$$ 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.

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.

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 $(V,\,F^\times\cdot h)$ 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 ${\rm PU}(V,h)$ correspond bijectively 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 $$1=H^1(F,K^\times)\to H^1(F, {\rm GU}(V,h))\to H^1(F,{\rm Aut}({\rm PU}(V,h))),$$ which shows that the kernel is trivial. For the second arrow we have 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.

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 $$1=H^1(F,K^\times)\to H^1(F, {\rm GU}(V,h))\to H^1(F,{\rm Aut}({\rm PU}(V,h))),$$ 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.

No. 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.

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 $(V,\,F^\times\cdot h)$ 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 ${\rm PU}(V,h)$ correspond bijectively 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 $$1=H^1(F,K^\times)\to H^1(F, {\rm GU}(V,h))\to H^1(F,{\rm Aut}({\rm PU}(V,h))),$$ which shows that the kernel is trivial. For the second arrow we have 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.