$$g\in Z_{G}(H\cap
\Ad(g)H)H.$$ By Lemma \[L:G2-2\], we have $H\cap\Ad(g)H=H^{g^{3}}$. Replacing $g$ by an $H$ conjugate element if necessary, we may assume that $g\in Z$ for $Z=Z(K')$ as in the proof of Lemma \[L:G2-1\].

When $g^6\neq 1$, we have $Z_{H}(g)=Z_{H}(g^{3})$ by Lemma \[L:G2\]. Thus, $$g\in Z_{G}(H^{3})\subset
Z_{G}(H\cap\Ad(g)H)H.$$ When $g^6=1$, we have $g^2\in Z\cap K\subset H$ and $g^3\in Z_{G}(H\cap\Ad(g)H)$. Thus, $$g=g^{3}g^{-2}\in Z_{G}(H\cap\Ad(g)H)H.$$

Pinned automorphisms
--------------------

[*The pair $\O(2n+1)\subset\SO(2n+2)$.*]{} Let $n\geq 1$, $G=\SO(2n+2)$ and $\theta=
\Ad(I_{2n+1,1})$. Then, $\theta^2=1$ and $G^{\theta}=\O(2n+1)$. Write $H=G^{\theta}$. Write $$g_{\theta}=\left(\begin{array}{ccc}I_{2n}&0&0\\0&\cos\theta&\sin\theta\\0&-\sin\theta&\cos\theta
\\\end{array}\right)$$ ($\theta\in\mathbb{R}$). By Lemma \[L:symmetric2\], we have $G=HAH$ for $$A=\{g_{\theta}:\theta\in[0,2\pi)\}.$$ We know that the condition $(\ref{Eq2})$ in Lemma \[L:SCF2\] fails for an element $g\in A$ if and only if it fails for all elements in the double coset $HgH$.

\[P:O(2n+1)-1\] For $g=g_{\theta}\in A$ ($\theta\in[0,2\pi)$), the condition $(\ref{Eq2})$ in Lemma \[L:SCF2\] fails if and only if $\theta=\frac{\pi}{2}$ or $\frac{3\pi}{2}$.

When $\theta\in[0,2\pi)-\{0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\}$, one could verify that $G^{g^2}=G^{g}$. Thus, $$g\in Z_{G}(H^{g})=Z_{G}(H^{g^2})\subset Z_{G}(H^{g^2})H.$$ Hence, the condition $(\ref{Eq2})$ in Lemma \[L:SCF2\] holds. When $\theta=0$ or $\pi$, $$g\in H\subset Z_{G}(H^{g^2})H.$$ Hence, the condition $(\ref{Eq2})$ in Lemma \[L:SCF2\] holds.

When $\theta=\frac{\pi}{2}$ or $\frac{3\pi}{2}$, $g^2=g_{\pi}=\diag\{I_{2n},-1,-1\}$. In this case $H^{g^{2}}=S(\O(2n)\!\times\!\O(1)\!\times\!\O(1))$. Thus, $Z_{G}(H^{g^2})\subset H$ and $Z_{G}(H^{g^2})H=H$. However, $g\not\in H$. Hence, the condition $(\ref{Eq2})$ in Lemma \[L:SCF2\] fails.

Different with 