I would like to see if the following statement is true in general.
Statement: Let $G$ be a topological group acting on a topological space $X$ continuously. Suppose that a sequence $g_n \in G$ and an element $x \in X$ satisfy $g_n \cdot x \to g \cdot x$ for some $g \in G$. Then we have $g_n^{-1} \cdot x \to g^{-1} \cdot x$.
Unfortunately, I do not know whether the statement is true or not.
Thank you.
It is not true, and in fact, you don't really even need to be taking a limit. Let $G=O_2$, the group of $2\times 2$ orthogonal matrices, acting on $\mathbb R^2$, and let $$g_n= \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\text{, } g = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\text{, } x=\begin{bmatrix} 1 \\ 1 \end{bmatrix}\text{,}$$ then $$g_n\cdot x= g\cdot x = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\text{,}$$ yet $$g_n^{-1}\cdot x = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\text{,}$$ whereas $$ g^{-1}\cdot x=\begin{bmatrix} -1 \\ 1 \end{bmatrix}\text{.}$$
If $G$ is compact Abelian and both $G$ and $X$ are metrizable, then it is true, as you can verify by passing to subsequences for which $\{g^{-1}\cdot g_n\}$ converges to a member of the stabilizer of $x$, whereby so does $\{g\cdot g_n^{-1}\}$.
You could probably relax those assumptions too, though I don't know how much.