This question is inspired by [On moments of inertia of planar and 3D convex bodies](https://mathoverflow.net/questions/435361/on-moments-of-inertia-of-planar-and-3d-convex-bodies).
Let $f:{\mathbb R}^3\setminus\{0\}\to{\mathbb R}$ be an even homogeneous ($f(kx)=f(x)$ for all real $k\neq 0$) continuous function.
Can one always find an orthogonal basis $(x\_1,x\_2,x\_3)$ such that
$f(x\_1)=f(x\_2)=f(x\_3)$ ?


Remark. Even without the condition that $f$ is even, I do not know a counterexample. This condition comes from the application that I mention in the first sentence.


Remark 2. Thanks to @Aleksei Kulikov, who found the reference: this was a problem posed by Rademacher and solved by Kakutani in 1942 (Annals of math., 43 (1942) 739-741). Most interestingly, the reviewer
in MR0007267 wrote that the corresponding problem in $R^n, n\geq 4$ was unsolved in 1942.
I found that the $n$-diensonal generalization was obtained by H. Yamabe and Z. Yujobô,
Osaka Math. J. 2 (1950), 19–22.



A property of even continuous functions on the sphere
It seems there aren't any counterexamples, even if $f$ is homogeneous but not even.


If there is some counterexample $f$ to the question, and letting $X=\mathbb{R}^3\setminus\{x\_1=x\_2=x\_3\}$, we can consider the map $F:SO(3)\to X$, given by $F(u,v,w)=(f(u),f(v),f(w))$ (we will sometimes represent elements $r\in SO(3)$ by $(r(e\_1),r(e\_2),r(e\_3))$, $(e\_i)\_{i=1}^3$ being the usual basis of $\mathbb{R}^3$).


As $\pi\_1(SO(3))\cong\mathbb{Z}\_2$ and $\pi\_1(X)\cong\mathbb{Z}$, we will have $F\_\*(\gamma)=0\;\forall\gamma\in\pi\_1(SO(3))$.


Now consider the loop $\gamma:t\mapsto\gamma\_t=r\_{2\pi t}(v)$ given by rotations of angle $2\pi t$ around $v:=(1,1,1)$. Note that $\gamma\_{\frac{1}{3}}(e\_1)=e\_2,\gamma\_{\frac{1}{3}}(e\_2)=e\_3$ and $\gamma\_{\frac{1}{3}}(e\_3)=e\_1$. So if $\gamma\_t=(u,v,w)$, then $\gamma\_{t+\frac{1}{3}}=(\gamma\_{t+\frac{1}{3}}(e\_1),\gamma\_{t+\frac{1}{3}}(e\_2),\gamma\_{t+\frac{1}{3}}(e\_3))=(\gamma\_{t}(e\_2),\gamma\_{t}(e\_3),\gamma\_{t}(e\_1))=(v,w,u)$.


So the path $\alpha=F\_\*\gamma\in\pi\_1(X)$ satisfies that if $\alpha(t)=(x\_1,x\_2,x\_3)$, then $\alpha(t+\frac{1}{3})=(x\_2,x\_3,x\_1)$. From this we will deduce that $\alpha$ is not trivial, which is a contradiction. To deduce that $\alpha$ is not trivial, we can first deformation retract $X$ to a circumference perpendicular to the line $\{x\_1=x\_2=x\_3\}$, so that identifying this circumference with $\mathbb{S}^1$, the rotation $(x\_1,x\_2,x\_3)\mapsto(x\_2,x\_3,x\_1)$ becomes the rotation $r\_{\frac{2\pi}{3}}$ of $\mathbb{S}^1$.


So we have to prove that any loop $\beta:[0,1]\to\mathbb{S}^1$ satisfying $\beta(t+\frac{1}{3})=e^{\frac{2\pi i}{3}}\beta(t)$ is not trivial. This is true because as $\beta(\frac{1}{3})=e^{\frac{2\pi i}{3}}$, the path $\beta|\_{[0,\frac{1}{3}]}$ rotates a total angle of $2\pi(n+\frac{1}{3})$ for some $n\in\mathbb{Z}$, so the winding number of $\beta$ is $3(n+\frac{1}{3})\neq0$.


