Let $(-\Delta_p)^s$ is the fractional $p$-Laplace operator defined in the usual principal value sense. In the local case, $s=1$, the p-Laplace operator is rotationally invariant under an orthogonal map. Does the same hold for the fractional p-Laplacian?
I meant the following definition:
$$ (-\Delta_p)^s u(x)=\lim_{\epsilon\to 0}\int_{\mathbb{R}^n\setminus B_{\epsilon}(x)}\frac{|u(x)-u(y)|^{p-2}(u(x)-u(y))}{|x-y|^{n+ps}}\,\,\mathrm{d}y, $$ where $0<s<1<p<\infty$.
As @F_M_ mentions in the comments, it is. Here is a proof that doesn't require the fourier transform. Let $\mathcal O$ be an orthogonal matrix and $v(x)=u(\mathcal O x)$. Then to show $(-\Delta_p)^s $ is rotationally invariant, we must prove that $(-\Delta_p)^s v(x) = (-\Delta_p)^su(\mathcal Ox)$.
By definition, \begin{align*} (-\Delta_p)^s v(x) &= \lim_{\varepsilon \to 0^+} \int_{\mathbb R^n \setminus B_\varepsilon (x)} \frac{\vert v(x)-v(y)\vert^{p-2} (v(x)-v(y))}{\vert x-y \vert^{n+ps}} \, dy \\ &= \lim_{\varepsilon \to 0^+} \int_{\mathbb R^n \setminus B_\varepsilon (x)} \frac{\vert u(\mathcal O x)-u(\mathcal O y)\vert^{p-2} (u(\mathcal O x)-u(\mathcal O y))}{\vert x-y \vert^{n+ps}} \, dy. \end{align*} Since $\mathcal O :\mathbb R^n \to \mathbb R^n$ is an isometry, $\vert x-y \vert= \vert\mathcal O x-\mathcal Oy \vert$, so $$(-\Delta_p)^s v(x) =\lim_{\varepsilon \to 0^+} \int_{\mathbb R^n \setminus B_\varepsilon (x)} \frac{\vert u(\mathcal O x)-u(\mathcal O y)\vert^{p-2} (u(\mathcal O x)-u(\mathcal O y))}{\vert\mathcal O x-\mathcal Oy \vert^{n+ps}} \, dy. $$ Making the change of variables $z=\mathcal O y $, noting that the Jacobian is $ \vert \partial y / \partial z\vert = \vert \det \mathcal O \vert =1 $, we have that \begin{align*} (-\Delta_p)^s v(x) &=\lim_{\varepsilon \to 0^+} \int_{\mathbb R^n \setminus B_\varepsilon (\mathcal O x)} \frac{\vert u(\mathcal O x)-u(\mathcal z)\vert^{p-2} (u(\mathcal O x)-u(\mathcal z))}{\vert\mathcal O x-\mathcal z \vert^{n+ps}} \, dz =(-\Delta_p)^s u(\mathcal O x) \end{align*} as required.