Let $\mathbb{S}^3_+$ be the set of $3 \times 3$ symmetric, positive semidefinite matrices. I wonder whether I can write $\mathbb{S}^3_+$ as a norm cone, i.e., $$\exists A\in \mathbb{R}^{m\times 9}, C, \|\cdot\|, s.t. X \in \mathbb{S}^n_+ \iff \|A\text{vec}(X)\|\leq \text{trace}(CX),$$
where $\text{vec}(X)$ means to put $X$ as a vector.
I guess not as there could be so many polynomial inequalities for the positive semidefinite case (the determinant of all the principal minors). But I could not figure out why. I have tried to look at the dual cone (as $\mathbb{S}_+^3$ is self-dual) but it seems very hard to compute the dual cone of $\{X:\|A\text{vec}(X)\|\leq \text{trace}(CX)\}$.
I am thinking whether there is any property always preserved by linear transform but not sure there is really any.
I'm going to assume that we may consider matrix norms. Then in fact, the answer is precisely analogous to the answer given for $\mathbb{R}^n_+$ in your other question.
Let $\|X\|_*$ be the nuclear norm of $X\in\mathbb{S}^n$; that is, $$\|X\|_*=\sum_{i=1}^n \sigma_i(X) = \sum_{i=1}^n|\lambda_i(X)|$$ Then we have $$X\in\mathbb{S}^n_+ \quad\Longrightarrow\quad \|X\|_* \leq \mathop{\textrm{Tr}}(X)$$ After all, this inequality is equivalent to $$\sum_{i=1}^n |\lambda_i(X)| \leq \sum_{i=1}^n \lambda_i(X)$$ and the only way for this to be true is if $\lambda_i(X)\geq 0$ for all $i$. Indeed, it holds with equality in that case.
One thing I should point out is that, in both cases, strict inequality cannot be satisfied; that is, there is no $X$ such that $\|X\|_*<\mathop{\textrm{Tr}}(X)$; or, in the vector case, no $x$ such that $\|x\|_1<\sum_i x_i$. This has some important practical consequences if, say, you intend to implement some sort of numerical method around this approach. Better to stick with the original conic form.