Kleiner' s Lemma and Slice Theorem Let $μ : G×M→M$ be an isometric action and let $γ : [0,1]→M$ be a
geodesic segment between G(γ(0)) and G(γ(1)) that realizes the distance between
these orbits. There exists a subgroup H of G such that $G_{γ(t)} = H$ for $t \in (0,1)$ and
$H$ is a subgroup of $G_{γ (0)}$ and $G_{γ (1)}$.
Suggestion of prove:
If $μ : G×M→M$ is an isometric action, então $\mu^g: M \longrightarrow M $ is a isometry, where $\mu^g(x)$ is the action of $g$ in $x$,i.e, $\mu^g(x)=\mu(g,x)$
Let
$H :=\{ g ∈ G : μ(g,γ (t)) = γ (t)), \space \forall  t \in [0,1]\}$. Suppose there exists
some $t_0 ∈ (0,1)$ and $g \in G_{γ(t_0)}$ such that $g \notin H$. In particular,
$$d(μ^g)_{γ (t_0)}(γ'(t_0))\neq γ'(t_0)$$.
Define a piecewise smooth path
$γ :[0,1]→M $ by setting $γ|_{[0,t_0]} := γ[0,t_0]$ and $γ|_{[t_0,1]} := μ(g,γ[t_0,1])$. Then
$\gamma$ joins $G(γ(0))$ to $G(γ(1))$ and has the same length as $γ$, contradicting the fact that minimizing geodesic segments are smooth.
my question:
I) how to justify the part of, in particular $$d(μ^g)_{γ (t_0)}(γ'(t_0))\neq γ'(t_0),$$ using Slice's Theorem?
II) And how to justify the application of the first variation formula in the final part.
 A: To answer I, I think you have a typo.  Namely, for $g\in G_{gamma(t_0)}$, I think $g\in H$ iff $d(\mu^g)_{\gamma(t_0)} \gamma'(t_0) = \gamma'(t_0)$.  This is a consequence of the fact that geodesics are determined by a single point and single tangent vector.
Thus, $\mu^g(\exp(t \gamma'(t_0))) = \exp(t (d(g^\mu)_{\gamma(t_0)} \gamma'(t_0)))$ since they agree to first order at $t=0$.  Moreover, $d(g^\mu)_{\gamma(t_0)} \gamma'(t_0) = \gamma'(t_0)$ iff $\exp(d(g^\mu)_{\gamma(t_0)} \gamma'(t_0)) = \exp( t\gamma'(t_0))$ for all $t$ for which either side is defined.   Putting these together, we see that $d(g^\mu)_{\gamma(t_0)} \gamma'(t_0) = \gamma'(t_0)$ iff $g\in H$.
To answer II, note that we just need to show that minimizing curves must be smooth.  This follows from the existence of totally normal neighborhoods (that is, an open set for which any two points are connected by a unique length minimizing geodesic).
Specifically, suppose $\alpha$ has a corner at $t=0$.  Pick a totally normal neighborhood $U$ of $\alpha(0)$.  Pick $\epsilon$ small enough so that $\alpha(\pm \epsilon)\in U$.  Because we are in a totally normal neighborhood, there is a unique minimizing geodesic $\beta$ connecting $\alpha(-\epsilon)$ to $\alpha(\epsilon)$.  Geodesics are $C^\infty$, so $\beta$ is $C^\infty$.  On the other hand $\alpha$ is not $C^\infty$ so $\alpha\neq \beta$.  Since $\beta$ is the unique length minimizer between $\alpha(-\epsilon)$ and $\alpha(\epsilon)$, it follows that $\alpha$ does not minimize distance.
