Consider the case $n=1$. Let $m$ be any natural number, and let $G=\mathbb{C}^{\times}$ act on $\mathbb{C}$ by $g\cdot v = g^mv$. Consider the sequence $g_k = e^{2\pi ik/m}$. Take $v=1$. Then $g_k\cdot v = v$ but the sequence $g_k$ is periodic and does not converge. The stabilizer here is the group of $m$-th roots of unity, which is finite.

However, there is always a subsequence which converges:

Since $H$ is finite there exists an $\epsilon>0$ such that the sets $h\cdot B(1,\epsilon)$ are disjoint, where $B(1,\epsilon)=\{x|d(1,x)\leq\epsilon\}$.

Consider the continuous map $f:G\to V$ which sends $g$ to $g\cdot v$. This is a polynomial map which is also locally injective (due to the finiteness of $H$). This already implies that this map is a local homeomorphism. It the follows that there is a $\delta>0$ such that $f^{-1}(B(v,\delta))\subseteq \cup_{h\in H} h\cdot B(1,\epsilon)$.

Assume now that $\{g_k\cdot v\}$ converges to $v$. It follows that for almost all $k$ we have that $g_k\in h\cdot B(1,\epsilon)$ for some $h\in H$. Since $H$ is finite, this implies that for some $h\in H$ there are infinitely many $k$'s such that $g_k\in h\cdot B(1,\epsilon)$. These infinitely many $k$'s gives us a subsequence which converges to $h\in H$.

Yes, there is always a subsequence which converges:

Since $H$ is finite there exists an $\epsilon>0$ such that the sets $h\cdot B(1,\epsilon)$ are disjoint, where $B(1,\epsilon)=\{x|d(1,x)\leq\epsilon\}$.

Consider the continuous map $f:G\to V$ which sends $g$ to $g\cdot v$. This is a polynomial map which is also locally injective (due to the finiteness of $H$). This already implies that this map is a local homeomorphism. It the follows that there is a $\delta>0$ such that $f^{-1}(B(v,\delta))\subseteq \cup_{h\in H} h\cdot B(1,\epsilon)$.

Assume now that $\{g_k\cdot v\}$ converges to $v$. It follows that for almost all $k$ we have that $g_k\in h\cdot B(1,\epsilon)$ for some $h\in H$. Since $H$ is finite, this implies that for some $h\in H$ there are infinitely many $k$'s such that $g_k\in h\cdot B(1,\epsilon)$. These infinitely many $k$'s gives us a subsequence which converges to $h\in H$.

Consider the case $n=1$. Let $m$ be any natural number, and let $G=\mathbb{C}^{\times}$ act on $\mathbb{C}$ by $g\cdot v = g^mv$. Consider the sequence $g_k = e^{2\pi ik/m}$. Take $v=1$. Then $g_k\cdot v = v$ but the sequence $g_k$ is periodic and does not converge. The stabilizer here is the group of $m$-th roots of unity, which is finite.

Consider the case $n=1$. Let $m$ be any natural number, and let $G=\mathbb{C}^{\times}$ act on $\mathbb{C}$ by $g\cdot v = g^mv$. Consider the sequence $g_k = e^{2\pi ik/m}$. Take $v=1$. Then $g_k\cdot v = v$ but the sequence $g_k$ is periodic and does not converge. The stabilizer here is the group of $m$-th roots of unity, which is finite.

However, there is always a subsequence which converges:

Since $H$ is finite there exists an $\epsilon>0$ such that the sets $h\cdot B(1,\epsilon)$ are disjoint, where $B(1,\epsilon)=\{x|d(1,x)\leq\epsilon\}$.

Consider the continuous map $f:G\to V$ which sends $g$ to $g\cdot v$. This is a polynomial map which is also locally injective (due to the finiteness of $H$). This already implies that this map is a local homeomorphism. It the follows that there is a $\delta>0$ such that $f^{-1}(B(v,\delta))\subseteq \cup_{h\in H} h\cdot B(1,\epsilon)$.

Assume now that $\{g_k\cdot v\}$ converges to $v$. It follows that for almost all $k$ we have that $g_k\in h\cdot B(1,\epsilon)$ for some $h\in H$. Since $H$ is finite, this implies that for some $h\in H$ there are infinitely many $k$'s such that $g_k\in h\cdot B(1,\epsilon)$. These infinitely many $k$'s gives us a subsequence which converges to $h\in H$.