If $X=S^1\cup \{(x, 0): x\in[-1, 1]\}$, then for every homeomorphism $F:X\to X$, $F((-1, 0))=(-1, 0)$. This implies that if $\varphi:G\times X\to X$ is a continuous action, then $\varphi(g, (-1, 0))= (-1, 0)$. Hence every continuous action $\varphi:G\times X\to X$ does have $p= (-1, 0)$ as fixed point.
But, referee of my manuscript claim that it is not true and give a continuous action on $X$ by rotations on $S^1$ and the identity map on $\{(x, 0): x\in(-1, 1)\}$.
Consider $\{ (-1-\frac{1}{n}, 0)\}_{n\in\mathbb{N}}$. Take $g\in G$, then $\{\varphi(g, (-1-\frac{1}{n}, 0))\}=\{(-1-\frac{1}{n}, 0)\} $ but if $\varphi(g, (-1, 0))\neq (-1, 0)$, then $(-1-\frac{1}{n}, 0)\to \varphi(g, (-1, 0))$ that is contradiction, because $(-1-\frac{1}{n}, 0)\to (-1, 0)$
Is it true every continuous action $\varphi:G\times X\to X$ does have $p= (-1, 0)$ as fixed point? please help me to know it.