Here is a proof of uniqueness. It is a bit long, but the best I have found. Any feedback is more than welcome.
Let $$C\equiv\{x\in\mathbb R^n\,|\,\|x\|=r\}$$ denote the surface of the $n$-dimensional sphere of radius $r$. Let $\mu$ and $\nu$ denote the $n$-dimensional Borel probability measures generated by $X$ and $Y$, respectively. The assumptions on $X$ and $Y$ imply that
\begin{align*}
\mu(C)=&\,1\tag{$\spadesuit$},\\
\mu(U^{-1}(E))=&\,\mu(E)\quad\forall E\in\mathscr B^n\tag{$\clubsuit$}
\end{align*}
for any orthogonal transformation $U:\mathbb R^n\to\mathbb R^n$, where $\mathscr B^n$ denotes the $n$-dimensional Borel $\sigma$-algebra on $\mathbb R^n$. The analogous properties hold for $\nu$. In what follows let $$B(\varepsilon,x)\equiv\{y\in\mathbb R^n\,|\,\|x-y\|<\varepsilon\}$$ denote the ball of radius $\varepsilon>0$ around $x\in\mathbb R^n$.
Claims 1–4 below only use the properties ($\spadesuit$)–($\clubsuit$), so they remain true if $\mu$ is replaced with $\nu$.
Claim 1: For any fixed $\varepsilon>0$, the balls $B(\varepsilon,x)$ have the same $\mu$-measure for any $x\in C$.
Proof: If $x,y\in C$, then there exists an orthogonal transformation $U:\mathbb R^n\to\mathbb R^n$ such that $U(x)=y$ (see the answers to another question of mine here). It follows that $U^{-1}(B(\varepsilon,y))=B(\varepsilon,x)$ and the claim follows from ($\clubsuit$). $\quad\blacksquare$
Claim 2: If $x\in C$ and $\varepsilon >0$, then $\mu(B(\varepsilon,x))>0$.
Proof: Since $C\subseteq\bigcup_{x\in C}B(\varepsilon,x)$ and $C$ is compact, there exist some $m\in\mathbb N$ and $\{x_1,\ldots,x_m\}\subseteq C$ such that $C\subseteq \bigcup_{i=1}^m B(\varepsilon,x_i)$. Therefore, $$1=\mu(C)\leq\sum_{i=1}^m\mu(B(\varepsilon,x_i)).$$ By Claim $1$, it follows that $$\mu(B(\varepsilon,x))\geq\frac{1}{m}>0$$ for any $x\in C$. $\quad\blacksquare$
Define a function $F:(0,\infty)\to(0,\infty)$ as $$F(\varepsilon)\equiv \mu(B(\varepsilon,x))\quad\text{for $\varepsilon>0$},$$ where $x\in C$ is arbitrary. By Claim 1, this function is well-defined and does not depend on the particular choice of $x\in C$, and by Claim 2, it assumes positive values. The corresponding function where $\mu$ is replaced with $\nu$ will be denoted as $G$.
Claim 3: Let $E\subseteq\mathbb R^n$ be any open set such that $E\cap C\neq\varnothing$. Then, $$\lim_{n\to\infty}\frac{\mu(E\cap B(n^{-1},x))}{F(n^{-1})}=1\quad\text{for any $x\in E\cap C$}.$$
Proof: If $x\in E\cap C$, then for $n\in\mathbb N$ large enough, $B(n^{-1},x)\subseteq E$. $\quad\blacksquare$
Claim 3 continues to hold if $\mu$ is replaced with $\nu$ and $F$ with $G$.
Claim 4: For any Borel set $E\in\mathscr B^n$ and $\varepsilon>0$, the indicator function $(x,y)\mapsto \mathbb I_{E\cap B(\varepsilon,x)}(y)$ from $\mathbb R^n\times\mathbb R^n$ to $\mathbb R$ is measurable.
Proof: Clearly, $(x,y)\mapsto \mathbb I_E(y)$ is measurable, so it is sufficient to show that $(x,y)\mapsto\mathbb I_{B(\varepsilon,x)}(y)$ is measurable. But the value of this function is $1$ if and only if $\|x-y\|<\varepsilon$, which holds on an open subset of $\mathbb R^n\times\mathbb R^n$. $\quad\blacksquare$
Note that Claim 4 implies that the function $$x\mapsto \mu(E\cap B(\varepsilon,x))=\int_{y\in\mathbb R^n}\mathbb I_{E\cap B(\varepsilon,x)}(y)\,\mathrm d\mu(y)$$
from $\mathbb R^n$ to $\mathbb R$ is measurable.
Now I will show that $\mu$ and $\nu$ agree on open sets, which will imply that they agree on all Borel sets, completing the proof of uniqueness. This argument is based on Theorem 3.4 in Mattila (1999, pp. 45–46). Let $E\subseteq\mathbb R^n$ be an open set. If $E\cap C=\varnothing$, then ($\spadesuit$) implies that $\mu(E)=\nu(E)=0$. Suppose, therefore, that $E\cap C\neq\varnothing$. Then, one has the following:
\begin{align*}
\nu(E)=&\,\nu(E\cap C)=\int_{x\in E\cap C}1\,\mathrm d\nu(x)=\int_{x\in E\cap C}\liminf_{n\to\infty}\left\{\frac{\mu(E\cap B(n^{-1},x))}{F(n^{-1})}\right\}\mathrm d\nu(x)\\
\leq&\,\liminf_{n\to\infty}\left\{\int_{x\in E\cap C}\frac{\mu(E\cap B(n^{-1},x))}{F(n^{-1})}\mathrm d\nu(x)\right\}\\=&\,\liminf_{n\to\infty}\left\{\frac{1}{F(n^{-1})}\int_{x\in E\cap C}\int_{y\in E\cap C}\mathbb I_{E}(y)\mathbb I_{B(n^{-1},x)}(y)\,\mathrm d\mu(y)\,\mathrm d\nu(x)\right\}\\
=&\,\liminf_{n\to\infty}\left\{\frac{1}{F(n^{-1})}\int_{y\in E\cap C}\int_{x\in E\cap C}\mathbb I_{B(n^{-1},y)}(x)\,\mathrm d\nu(x)\,\mathrm d\mu(y)\right\}\\
\leq&\,\liminf_{n\to\infty}\left\{\frac{1}{F(n^{-1})}\int_{y\in E\cap C}\int_{x\in\mathbb R}\mathbb I_{B(n^{-1},y)}(x)\,\mathrm d\nu(x)\,\mathrm d\mu(y)\right\}\\
=&\,\liminf_{n\to\infty}\left\{\frac{1}{F(n^{-1})}\int_{y\in E\cap C}\nu(B(n^{-1},y))\,\mathrm d\mu(y)\right\}\\
=&\,\liminf_{n\to\infty}\left\{\frac{G(n^{-1})}{F(n^{-1})}\mu(E)\right\}\\
=&\,\liminf_{n\to\infty}\left\{\frac{G(n^{-1})}{F(n^{-1})}\right\}\mu(E).
\end{align*}
where I used Claims 1 and 3, Fatou’s lemma, Tonelli’s theorem, and measurability of the integrands follows from Claim 4. A symmetric argument reveals that $$\mu(E)\leq\liminf_{n\to\infty}\left\{\frac{F(n^{-1})}{G(n^{-1})}\right\}\nu(E).$$ In particular, if one were to choose $E\equiv B(m^{-1},x)$ for some $x\in C$ and $m\in\mathbb N$, then the first of the two inequalities just derived implies (the division is legitimiate–see Claim 2) that $$\frac{G(m^{-1})}{F(m^{-1})}=\frac{\nu(E)}{\mu(E)}\leq\liminf_{n\to\infty}\left\{\frac{G(n^{-1})}{F(n^{-1})}\right\}.$$ Since this true for all $m\in\mathbb N$, it follows that $$\limsup_{n\to\infty}\left\{\frac{G(n^{-1})}{F(n^{-1})}\right\}\leq \liminf_{n\to\infty}\left\{\frac{G(n^{-1})}{F(n^{-1})}\right\},$$ so that $\lim_{n\to\infty}\{G(n^{-1})\div F(n^{-1})\}$ exists (and is possibly infinite). Similarly, $\lim_{n\to\infty}\{F(n^{-1})\div G(n^{-1})\}$ exists as well (in $\mathbb R\cup\{+\infty\}$).
Now, if either one of these two limits is infinite, then the other is $0$ and the two key inequalities on $\mu(E)$ and $\nu(E)$ imply that $\mu(E)=\nu(E)=0$, which is impossible if, say, $E=\mathbb R^n$. If both of these limits are finite and one of them vanishes, then the other limit is infinite, and it follows that $\mu(E)=\nu(E)=0$, which is, again, impossible. Therefore, both limits must be finite and positive, so that they are reciprocals of each other. Consequently
\begin{align*}
\mu(E)\leq&\,c\nu(E),\\
\nu(E)\leq&\,c^{-1}\mu(E)
\end{align*}
for some $c>0$ (universal across any open $E\subseteq \mathbb R^n$ meeting $C$). Letting $E=\mathbb R^n$, it immediately follows that $c=1$, which implies that $\mu(E)=\nu(E)$ for general $E\subseteq\mathbb R^n$ that is open and meets $C$. The proof is now complete.
