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Being the question very ample, its ampleness reflects itself in the possible answers. So, even if trivial, I give a basic observation just in order to compose the complessive frame:

A linear differential operator L on the Schwartz space in $\mathbb{R}^n$ is invariant under the euclidean group, i.e. $L\circ {R^{}}={R^{}}\circ R^{\ast}}={R^{\ast}}\circ L$, for any $R\in E(n)$, if and only if there is a polynomial in one indeterminate $P(T)$ such that $L=P(\Delta)$.

For a reference see Theorem 8.51 in Folland G., Real Analysis, 2nd Ed..

Excuse me if this observation is too elementary.
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Being the question very ample, its ampleness reflects itself in the possible answers. So, even if trivial, I give a basic observation just in order to compose the complessive frame:

A linear differential operator L on the Schwartz space in $R^n$ \mathbb{R}^n$ is invariant under the euclidean group, i.e. $L\circ R^=R^\circ {R^{}}={R^{}}\circ L$, for any $R\in E(n)$, if and only if there is a polynomial in one indeterminate $P(T)$ such that $L=P(\Delta)$.

For a reference see : Theorem 8.51 in Folland G., Real Analysis, 2nd Ed., Theorem 8.51Ed..

Excuse me if this observation is too elementary.
show/hide this revision's text 1

Being the question very ample, its ampleness reflects itself in the possible answers. So, even if trivial, I give a basic observation just in order to compose the complessive frame:

A linear differential operator L on the Schwartz space in $R^n$ is invariant under the euclidean group, i.e. $L\circ R^=R^\circ L$, for any $R\in E(n)$, if and only if there is a polynomial in one indeterminate $P(T)$ such that $L=P(\Delta)$.

For a reference see: Folland G., Real Analysis, 2nd Ed., Theorem 8.51.