Let $g$ be a complete $SO(n)$-invariant metric on $\mathbb R^n$. That is, for each $A \in SO(n)$, the map $\varphi_A:\mathbb R^n \rightarrow \mathbb R^n$ given by $x \mapsto Ax$ is a Riemannian isometry of $(\mathbb R^n,g)$.
Let $\exp_0:T_0\mathbb R^n \rightarrow \mathbb R^n$ denote the Riemannian exponential map at zero. Recall that $\exp_0(v) := \gamma_v(1)$, where $\gamma_v:\mathbb R \rightarrow \mathbb R^n$ is the unique maximal geodesic starting at zero with initial velocity $v$. Since $g$ is complete, $\exp_0$ is surjective (there exists a geodesic connecting any point in $\mathbb R^n$ to zero). It is well-known that $\exp_0$ is a diffeomorphism if restricted to a sufficiently small neighbourhood around zero.
Question: Is $\exp_0$ a diffeomorphism?
(I know that it remains to show $\exp_0$ is injective and a local diffeomorphism, but I'm not sure how to show these.)
EDIT:
I believe the answer is yes and I have come up with a solution. Please let me know if I have made a mistake somewhere.
Consider the inner product space $(T_0 \mathbb R^n,g_0)$. Observe that the isotropy subgroup of $0 \in \mathbb R^n$ is the entire group $SO(n)$. Thus, $SO(n)$ acts by linear isometries on $(T_0 \mathbb R^n,g_0)$ via $A \cdot v := d \varphi_A(v)$. For each $r > 0$, let us define $$S_r(0) := \left\{v \in T_0 \mathbb R^n \mid g_0(v,v) = r^2\right\}.$$
Lemma 1. For any $r > 0$, $SO(n)$ acts transitively on $S_r(0)$.
Proof. First, let us identify $T_0 \mathbb R^n$ with $\mathbb R^n$ via the linear isomorphism $\partial_i|_0 \mapsto e_i$. Here, $\partial_i|_0$ are the coordinate vectors induced by the identity chart, and $e_i$ denote the standard basis for $\mathbb R^n$. Recall that under this identification, for each $A \in SO(n)$, the differential $d\varphi_A|_0:T_0 \mathbb R^n \rightarrow T_0\mathbb R^n$ is just $\varphi_A:\mathbb R^n \rightarrow \mathbb R^n$, because $\varphi_A$ is linear. Thus, under this identification, the action of $SO(n)$ on $T_0 \mathbb R^n \cong \mathbb R^n$ is exactly the same as the original action.
We know that $SO(n)$ acts transitively on the Euclidean spheres $\mathbb S^{n-1}(R)$. Thus, we are done if we show that $g_0$ is a multiple of the standard dot product $\bullet$ on $\mathbb R^n$. Set $c := g_0(e_1,e_1)$. Fix an arbitrary element in $\mathbb R^n$, which can be written as $tv$, with $t \geq 0$ and $v\bullet v = 1$. Since $SO(n)$ acts transitively on $\mathbb S^{n-1}(1)$, there exists $A \in SO(n)$ such that $Av = e_1$. Therefore, since each $d\varphi_A|_0=\varphi_A$ is a linear isometry of $(T_0 \mathbb R^n\cong \mathbb R^n,g_0)$, we find $$g_0(tv,tv) = g_0(Atv, Atv) = t^2 g_0(Av,Av) = t^2 g_0(e_1,e_1) = t^2 c = c (tv \bullet tv),$$ as desired. $\square$
Lemma 2. Fix $R > 0$. Then there exists some $r > 0$ such that the restriction $\exp_0|_{S_r(0)}:S_r(0) \rightarrow \mathbb S^{n-1}(R)$ is a surjective smooth submersion.
Proof. Since $\exp_0$ is surjective, there exists a unit speed geodesic $\gamma:\mathbb R \rightarrow \mathbb R^n$ such that $\gamma(0) = 0$ and $\gamma(r) \in \mathbb S^{n-1}(R)$ for some $r > 0$. Let $v := \gamma'(0)$.
First, let us show that the image of $S_r(0)$ is contained in $S^{n-1}(R)$. To this end,fix an arbitrary element of $S_r(0)$, which can be written as $rw$, where $r > 0$ and $w \in S_1(0)$. We wish to show that $\exp_0(rw) \in \mathbb S^{n-1}(R)$. By the previous lemma, we know that $SO(n)$ acts transitively on $S_1(0)$, so there exists some $A \in SO(n)$ such that $w = d \varphi_A(v) = (\varphi_A \circ \gamma)'(0).$ Uniqueness of geodesics implies that the geodesic $\varphi_A \circ \gamma$ is given by $t \mapsto \exp_0(tw)$. In particular, $A \gamma(r) = (\varphi_A \circ \gamma)(r) = \exp_0(rw)$. Thus, $\exp_0(rw)$ belongs to $\mathbb S^{n-1}(R)$.
Next, let us show surjectivity. Suppose $p \in \mathbb S^{n-1}(R)$. Then there exists $A \in SO(n)$ such that $p = A\gamma(r) = (\varphi_A \circ \gamma)(r)$. We know that $\varphi_A \circ \gamma$ is a unit speed geodesic starting at zero. Set $w := (\varphi_A \circ \gamma)'(0)$. By uniqueness of geodesics, we know that $\varphi_A \circ \gamma$ is given by $t \mapsto \exp_0(tw)$. In particular, $\exp_0(rw) = (\varphi_A \circ \gamma)(r) = p$.
Now, let us show that $\exp_0|_{S_r(0)}:S_r(0) \rightarrow \mathbb S^{n-1}(R)$ is a smooth submersion. Since we have shown surjectivity, the Global Rank Theorem tells us that it suffices to show that $\exp_0|_{S_r(0)}$ has constant rank. By Lemma 1, we know that $SO(n)$ acts transitively on the domain $S_r(0)$. Thus, by the Equivariant Rank Theorem, it suffices to show that $\exp_0|_{S_r(0)}:S_r(0) \rightarrow \mathbb S^{n-1}(R)$ is $SO(n)$-equivariant. Fix $rv \in S_r(0)$, where $r > 0$ and $v \in S_1(0)$. Then $$A\cdot \exp_0(rv) = (\varphi_A \circ \exp_0)(rv) = (\exp_0 \circ d \varphi_A |_0)(rv) = \exp_0(A \cdot rv),$$ where we have used the naturality of the exponential map in the second equality. $\square$
Lemma 3. Fix $R > 0$. Let $\gamma: \mathbb R \rightarrow \mathbb R^n$ be a unit speed geodesic such that $\gamma(0) = 0$ and $\gamma(r) \in \mathbb S^{n-1}(R)$. Then $\gamma'(r)$ is orthogonal to $T_{\gamma(r)}S^{n-1}(R)$.
Proof. Let $v := \gamma'(0)$. By Lemma 2, we know that $$d (\exp_0|_{S_r(0)})|_{rv}: T_{rv} S_r(0) \rightarrow T_{\gamma(r)} \mathbb S^{n-1}(R)$$ is a linear isomorphism. Let $w \in T_{\gamma(r)} \mathbb S^{n-1}(R)$. Then there exists $u \in T_{rv} S_r(0)$ such that $dexp_0|_{rv}(u) = w$. The Gauss Lemma tells us that $$g_{\gamma(r)}(w, \gamma'(r)) = g_{\gamma(r)}(d\exp_0|_{rv}(u),d\exp_0|_{rv}(v) ) = g_0(u,v) = 0.$$ This completes the proof. $\square$
Lemma 4. Let $\gamma: \mathbb R \rightarrow \mathbb R^n$ be a geodesic such that $\gamma(0) = 0$. Then $\gamma(r) \neq 0$ for all $r > 0$.
Proof. For the sake of contradiction, suppose that $\gamma(r) = 0$ for some $r > 0$. Since $[0,r]$ is compact and $\gamma$ is continuous, we know that the image $\gamma([0,r])$ is bounded in the Euclidean norm $\| \cdot \|$. Thus, there exists $R > 0$ such that $\|\gamma(t) \| < R$ for all $t \in [0,r]$.
Now, fix $p \in \mathbb S^{n-1}(R)$. Since $(\mathbb R^n,g)$ is complete, there exists a unit speed geodesic $\alpha: \mathbb R \rightarrow \mathbb R^n$ such that $\alpha(0) = 0$, $\alpha(s) = p$ for some $s > 0$, and $\alpha|_{[0,s]}$ is distance minimising with respect to $g$. Choose $A \in SO(n)$ so that $\gamma'(0) = d\varphi_A(\alpha'(0)) = (\varphi_A \circ \alpha)'(0)$. By uniqueness of geodesics, we know that $\gamma = \varphi_A \circ \alpha$. Therefore, $\gamma|_{[0,s]}$ minimises the distance (with respect to the metric $g$) between $0$ and $\gamma(s) \in \mathbb S^{n-1}(R)$. It follows that $r < s$.
Consider the reparameterisation $\beta: \mathbb R \rightarrow \mathbb R^n$ given by $\beta(t) := \gamma(t + r)$. Then $\beta$ is a unit speed geodesic such that $\beta(0) = 0$ and $\beta(s-r) = \gamma(s-r + r) = \gamma(s)$. However, observe that $$L_g(\beta|_{[0,s-r]}) = s-r < s = L_g(\gamma|_{[0,s]}),$$ so this contradicts the fact that $\gamma$ is distance minimising. $\square$
Proposition. The exponential map $\exp_0:T_0 \mathbb R^n \rightarrow \mathbb R^n$ is injective.
Proof. For the sake of contradiction, suppose that $\exp_0(rv) = \exp_0(sw)$ for some $r,s \geq 0$ and $v,w \in S_1(0)$. Let $\alpha,\beta:\mathbb R \rightarrow \mathbb R^n$ denote the unit speed geodesics given by $\alpha:t \mapsto \exp_0(tv)$ and $\beta:t \mapsto \exp_0(tw)$. Then $\alpha(r) = \beta(s)$. If $\alpha(r) = \beta(s) = 0$, then Lemma 4 implies that $r = s$, so $rv = sw$.
Thus, let us suppose $\alpha(r) =\beta(s)$ is non-zero. Then $r,s > 0$. Fix $R>0$ such that $\alpha(r) = \beta(s)$ belongs to $\mathbb S^{n-1}(R)$. By Lemma 3, we know that $\alpha'(r)$ and $\beta'(s)$ are both orthogonal to $T_{\alpha(r)}\mathbb S^{n-1}(R)$. Since $\mathbb S^{n-1}(R)$ is a codimension one embedded submanifold, we conclude that either $\alpha'(r) = \beta'(s)$ or $\alpha'(r) = - \beta'(s)$. Let us show that in $rv = sw$ in the first case, and the second case leads to contradiction.
Now, suppose $\alpha'(r) = \beta'(s)$. Without loss of generality, suppose $r \leq s$. Define $\gamma: \mathbb R \rightarrow \mathbb R^n$ to be the unit speed geodesic given by $\gamma(t) := \beta(t + s - r)$. Now, observe that $\gamma(r) = \beta(s) = \alpha(r)$ and $\gamma'(r) = \beta'(s) = \alpha'(r)$. Uniqueness of geodesics implies that $\gamma = \alpha$. In particular, $0 = \beta(0) = \gamma(s-r) = \alpha(s-r)$. Thus, by Lemma 4, we must have $s = r$. Therefore, $\alpha = \gamma = \beta$. In particular, $v = \alpha'(0) = \beta'(0) = w$, so $sv = rw$, as desired.
Finally, suppose that $\alpha'(r) = -\beta'(s)$. Let $\gamma: \mathbb R \rightarrow \mathbb R^n$ be the unit speed geodesic given by $\gamma(t) := \beta(-t + s + r)$. Then observe that $\gamma(r) = \beta(-r + s + r) = \beta(s) = \alpha(r)$ and $\gamma'(r) = - \beta'(-r + s + r) = -\beta'(s) = \alpha'(r)$, so uniqueness of geodesics implies that $\gamma = \alpha$. In particular, $\alpha(s + r) = 0$. However, Lemma 4 implies that this is a contradiction, since $s + r > 0$. $\square$
Proposition. The exponential map $\exp_0:T_0 \mathbb R^n \rightarrow \mathbb R^n$ is a diffeomorphism.
Proof. We have shown that $\exp_0:T_0 \mathbb R^n \rightarrow \mathbb R^n$ is bijective. By the Global Rank Theorem, it suffices to show that $\exp_0$ has constant rank. Thus, fix $r > 0$ and $v \in S_1(0)$, and let $\gamma: \mathbb R \rightarrow \mathbb R^n$ denote the unit speed geodesic given by $\gamma(t) := \exp_0(tv)$. Let $R > 0$ be such that $\gamma(r) \in \mathbb S^{n-1}(R)$. Consider the differential $d\exp_0|_{rv}:T_0 \mathbb R^n \rightarrow T_{\gamma(r)} \mathbb R^n$.
Let $e_2,\ldots,e_n$ denote unit vectors in $T_0 \mathbb R^n$ such that $\gamma'(0),e_2,\ldots,e_n$ forms an orthonormal basis for $T_0 \mathbb R^n$. Then we know that $e_2,\ldots,e_n$ must belong to $T_{rv} S_r(0)$. By Lemma 2, we know that $\exp_0|_{S_r(0)}:S_r(0) \rightarrow \mathbb S^{n-1}(R)$ is a submersion, so in particular, $d \exp_0|_{rv}(e_2),\ldots,d \exp_0|_{rv}(e_n)$ forms a basis of $T_{\gamma(r)} \mathbb S^{n-1}(R)$. Since $\gamma'(r) = d \exp_0|_{v}$ is orthogonal to these vectors, it follows that $d\exp_0|_{rv}$ is a linear isomorphism. This completes the proof. $\square$