Construction of certain equivariant map isotopic to identity. Suppose $M$ is a compact manifold equipped with group action of $G$.This action has finite isolated fixed points. The textbook then make the following statement.

We can construct an equivariant smooth map $g : M \rightarrow M$ which is isotopic to the identity and maps a neighborhood of the fixed point set onto the fixed point set.(For example, use the exponential map with respect to an invariant metric.)

I wonder how this construction exactly works out for the exponential map. Thanks for any comment. 
 A: Since the action of $G$ is with isolated fixed points, we can basically assume it has just one fixed point $p \in M$.
As suggested, take $g$ a $G$-invariant metric on $M$, i.e., if $l_g$ denotes the action with $g \in G$, then $g(l_{g,*} v , l_{g,*} w) = g(v,w)$ for any $v, w$ tangent vectors.
We can restrict $\exp_p$ to a ball around $0$, $U \subseteq T_pM$, such that it $\exp_p|_U$ is a diffeomorphism.
Then it's easy to see (using the definition of geodesics as minimizing the lagrangian defined by $g$ and the definition of $\exp$) that for any $g \in G$ and $v \in U$ we have $\exp_p (l_{g, *} (v)) = l_g( \exp_p(v))$; this makes sense because $p$ is fixed, so $l_{g, *} (v)$ stays in $T_p M$ and $l_{g,*}$ is an isometry, so $l_{g, *} (v)$ stays in $U$ (since $U$ was assumed to be a ball). This shows that the map we'll construct is equivariant.
Now just split your ball $U$ into three mutually disjoint pieces: a small disc $D$ around $0$ and two annuluses $A_1$ (outermost) and $A_2$ (innermost and around $D$) (a picture really helps here; it's really just 3 concentric circles, the outermost one the boundary of $U$) and take a function $f:U \rightarrow U$ which:

*

*is the identity on $A_1$

*sends all of $D$ to $0$

*dilates $A_2$ radially around $0$ such that it fills $D$ and leaves the common boundary with $A_1$ intact; so really the closer you get to $D$ in $A_2$, the more that point will get stretched towards $0$ on the same radius;

Now just take $\exp_p \circ f$ and extend this to the whole of $M$ by taking the identity outside of it. Since $f$ is homotopic to the identity, when composing with $\exp_p$ and extending like this you'll also obtain something homotopic to the identity. The equivariance is also clear since $l_{g, *}$ are isometries on $T_pM$ and $f$ only modifies the magnitudes.
