\[T:Griess\] (1) and Proposition \[P:SCF\], $\Sp(n)$ and $\O(n)$ are strongly acceptable as $\U(n)$ is. By Propositions \[P:SO(2n+1)\] and \[P:SCF\], so is $\SO(2n+1)$.

By Theorem \[T:Griess\] (2) and Proposition \[P:SCF\], $\G_2$ is strongly acceptable as $\SO(7)$ is.

Take an involution $\theta\in\G_2$. Then, $$\G_{2}^{\theta}\cong\Sp(1)^{2}/\langle(-1,-1)\rangle$$ (cf. [@Huang-Yu Table2]). By Proposition \[P:centralizer2\], $\SO(4)\cong\Sp(1)^{2}/\langle(-1,-1)$ is strongly acceptable as $\G_2$ is.

Examples of unacceptable compact Lie groups
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In the following, we construct a concrete example of two element-conjugate homomorphisms from $(C_4)^{2}$ which are not globally conjugate. This construction not only shows that $\SU(4)/\langle
-I\rangle$ (and also $\SO(6)$) is unacceptable, but also is the building block in showing many connected compact semisimple Lie groups are unacceptable in Theorem \[T:unacceptable1\] and Theorem \[T:unacceptable3\].

\[E:SU4\] Let $G=\SU(4)/\langle-I\rangle$, and $\Gamma=(C_4)^{2}$ with two generators $\gamma_1,\gamma_2$. Define $\phi:\Gamma\rightarrow\SU(4)$ by $$\phi(\gamma_1)=\diag\{1,1,\mathbf{i},-\mathbf{i}\},
\quad\phi(\gamma_2)=\diag\{1,\mathbf{i},1,-\mathbf{i}\}.$$ Define $\phi':\Gamma\rightarrow\SU(4)$ by $$\phi'(\gamma)=\overline{\phi(\gamma)},\ \forall\gamma\in\Gamma.$$ Write $\pi:\SU(4)\rightarrow
\SU(4)/\langle-I\rangle$ for the projection. Set $\rho=\pi\circ\phi$, and $\rho'=\pi\circ\phi'$. In the below, we show that $\rho$ and $\rho'$ are element-conjugate, but not globally conjugate. Thus, $\SU(4)/\langle-I\rangle$ is unacceptable.

Write $\gamma=\gamma_1^{a}\gamma_2^{b}$ ($a,b\in\{1,2,3,4\}$) for a general element in $\Gamma$. Then, $$\gamma(\gamma)=\diag\{1,\mathbf{i}^{b},\mathbf{i}^{a},(-\mathbf{i})^{a+b}\}.$$ When $(a,b)
\equiv(0,1),(1,0),(0,3),(3,0),(1,3),(3,1)$, $$\phi'(\gamma)=\overline{\phi(\gamma)}\sim\phi(\gamma);$$ when $(a,b)\equiv(2,1),(1,2),(2,3),(3,2),(1,1),(3,3)$, $$\phi'(\gamma)=\overline{\phi(\gamma)}\sim
-\phi(\gamma);$$ when $a$ and $b$ are both even, $$\phi'(\gamma)=\overline{\phi(\gamma)}=\phi(\gamma).$$ In any case we have $\rho'(\gamma)\sim\rho(\gamma)$. Thus, $\rho$ and 