Let $(M,\langle\cdot,\cdot\rangle)$ be a Riemannian manifold and $G$ be a Lie group action on $M$ by isometries.
Fix a point $p$ and take a vector $X$ at $T_pM$. Denote by $G_p$ the isotropy group at $p$. The group $G_p$ also acts on $M$ by isometries and by taking its differential we can furnish an $G_p$-action on $T_pM$. Therefore, we can define $$G_X = \{g \in G_p :\text{the differential of the isotropy action fix} ~X\}.$$
Let $\gamma(s) = \exp_p(sX), ~ s >0$ be a geodesic. Is it true that for every $s > 0$ we have $G_X \subset G_{\gamma(s)}?$
I think the answer to your question is yes.
If $g \in G_X$, we want to prove that $\alpha(s) = \gamma(s)$, where $\alpha(s) = g \cdot \gamma(s)$ for $s\geq 0$. Since $\gamma$ is a geodesic and translation by $g$ is an isometry, by hypothesis, we have that $\alpha$ is a geodesic. Clearly $\alpha(0) = p = \gamma(0)$, since $g \in G_p$ for starters.
Now we only have to check that $\alpha'(0) = \gamma'(0)$ and use uniqueness of geodesics. Ok, on one hand we have $\gamma'(0) = X$, and on the other hand $$\alpha'(0) = {\rm d}g_{p}(\gamma'(0)) = {\rm d}g_p(X) = X,$$since $g \in G_X$.