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As an answer to (1), consider any compact subgroup $G\subset O(n)$ and look at the $G$-invariant functions on $\mathbb{R}^n$ that are $G$-invariant and have homogeneity $1$. As long as $G$ does not act transitively on the space of lines in $\mathbb{R}^n$, that set of functions will be properly larger than the $O(n)$-invariant functions of homogeneity $1$ 1$, and any one of them that is sufficiently near the standard norm will be a norm. As a specific example, consider the irreducible action of$SO(3)$on the$5$-dimensional vector space$S$of traceless symmetric$3$-by-$3$matrices. There are two$SO(3)$-invariant polynomials on$S$, S$: $p_2(m) = trace(m^2)$ and $p_3(m) = det(m)$, and they . (They generate the ring of $SO(3)$-invariant polynomials on $S$. S$.) Now let $$|m| = \bigl(p_2(m)^3 + \epsilon\ p_3(m)^2\bigr)^{1/6}.$$ for some small$\epsilon$. When$\epsilon>0$is sufficiently small, this will be a strictly convex norm on$S$, and its symmetry group will be$SO(3)$, which acts irreducibly on$S$. 1 As an answer to (1), consider any compact subgroup$G\subset O(n)$and look at the$G$-invariant functions on$\mathbb{R}^n$that are$G$-invariant and have homogeneity$1$. As long as$G$does not act transitively on the space of lines in$\mathbb{R}^n$, that set of functions will be properly larger than the$O(n)$-invariant functions of homogeneity$1$and any one of them that is sufficiently near the standard norm will be a norm. As a specific example, consider the irreducible action of$SO(3)$on the$5$-dimensional vector space$S$of traceless symmetric$3$-by-$3$matrices. There are two$SO(3)$-invariant polynomials on$S$,$p_2(m) = trace(m^2)$and$p_3(m) = det(m)$, and they generate the ring of$SO(3)$-invariant polynomials on$S$. Now let $$|m| = \bigl(p_2(m)^3 + \epsilon\ p_3(m)^2\bigr)^{1/6}.$$ for some small$\epsilon$. When$\epsilon>0$is sufficiently small, this will be a strictly convex norm on$S$, and its symmetry group will be$SO(3)$, which acts irreducibly on$S\$.