Here is a different characterization of the subgroup $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)\subset\mathrm{SO}(n^2{-}1)$ that works when $n>2$.

Define a skew-symmetric trilinear form $\kappa:{\frak{su}}(n)\times{\frak{su}}(n)\times{\frak{su}}(n)\to\mathbb{R}$ by the formula
$$
\kappa(x,y,z)=\mathrm{tr}(xyz-xzy)\ \bigl(=\mathrm{tr}(yzx-yxz)=\mathrm{tr}(zxy-zyx)\bigr).
$$
Then, for $n>2$, the subgroup $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)\subset\mathrm{GL}\bigl({\frak{su}}(n)\bigr)$
is the identity component of the subgroup of $\mathrm{GL}\bigl({\frak{su}}(n)\bigr)$ that preserves $\kappa$. (This doesn't work for $n=2$, of course. Also, you do have to restrict to the identity component because $\mathrm{SU}(n)$ has an outer automorphism (namely, conjugation) when $n>2$, and this preserves $\kappa$ as well; accordingly, the full $\kappa$-stabilizer has two components.)

In a certain sense, this is the 'simplest' characterization of $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ as a subgroup of $\mathrm{GL}\bigl({\frak{su}}(n)\bigr)$. What I mean is that, if you want to describe this subgroup as the set of linear transformations that preserve some set of algebraic objects on the vector space ${\frak{su}}(n)$, then $\kappa$ is the object that sits in the smallest representation of $\mathrm{GL}\bigl({\frak{su}}(n)\bigr)$ that does the job.

For example, if you want to define $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ as the subgroup that preserves the Lie bracket $[,]:{\frak{su}}(n)\times {\frak{su}}(n)\to {\frak{su}}(n)$, then you have to think of $[,]$ as an element of ${\frak{su}}(n)\otimes {\frak{su}}(n)^\ast\otimes {\frak{su}}(n)^\ast$ or, if you want to build in the skew-symmetry of the bracket, as an element of ${\frak{su}}(n)\otimes \Lambda^2({\frak{su}}(n)^\ast)$. Either way, these vector spaces have a much higher dimension than $\Lambda^3({\frak{su}}(n)^\ast)$, which is where $\kappa$ lives.

As another example, if you want to think of $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ as the subgroup that preserves the quadratic form $Q(x) = \mathrm{tr}(x^2)$ and the ($\mathbb{R}$-valued) cubic form $C(x) = i\,\mathrm{tr}(x^3)$, you are asking that it preserve an element of $S^2({\frak{su}}(n)^\ast)\oplus S^3({\frak{su}}(n)^\ast)$, and, again, this has considerably higher dimension than $\Lambda^3({\frak{su}}(n)^\ast)$. You can cut this down a little bit by noting that $C$ is harmonic with respect to $Q$, so that you could look at characterizing $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ as the subgroup of $\mathrm{O}\bigl({\frak{su}}(n),Q\bigr)$ that stabilizes $C\in S^3_0({\frak{su}}(n)^\ast,Q)$, but this still is more total equations than the equations that say that an element fixes $\kappa$.

I'm *not* saying that using $\kappa$ to define $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ is the most useful way to define it, I'm just saying that it's algebraically the least redundant.

For example, look at the case $n=3$: The codimension of $\mathrm{Ad}\bigl(\mathrm{SU}(3)\bigr)$ in $\mathrm{GL}\bigl({\frak{su}}(3)\bigr)\simeq \mathrm{GL}(8,\mathbb{R})$ is $64-8 = 56$, which is equal to the dimension of $\Lambda^3\bigl(\mathbb{R}^8\bigr)$, so it follows that the $\mathrm{GL}(8,\mathbb{R})$-orbit of $\kappa$ is actually open in $\Lambda^3\bigl(\mathbb{R}^8\bigr)$. Thus, the condition of fixing $\kappa$ is $56$ independent equations on an element of $\mathrm{GL}(8,\mathbb{R})$, which is the absolute minimum number of equations you need to cut out $\mathrm{Ad}\bigl(\mathrm{SU}(3)\bigr)$ as a submanifold.