It is a well known fact that $SU(2)$ acts transitively on spheres inside its Lie algebra. My question is essentially this: does the same hold in the finite case, i.e. for $SU(2,q^2)$? I will make my question a little more precise.
We start with a finite field $F_q$ with $q = p^r$ elements, where $p$ is prime. $F_q$ has a unique (up to automorphism) quadratic field extension, which is isomorphic to $F_{q^2}$. Now, $F_{q^2}$ has an automorphism of order $2$, which we will denote by $\sigma$, which maps $x \to x^q$. Let $V$ be a $2$-dimensional vector space over $F_{q^2}$. We define an $F_q$-bilinear mapping $(-,-): V\times V \to F_{q^2}$ by
$(v,w) = \sum_{i=1}^2 v_i w_i^q$
The group of all $F_{q^2}$-linear automorphisms of $V$ which preserve $(-,-)$ is denoted by $U(2,q^2)$ and we denote its subgroup consisting of linear maps with determinant equal to $1$ by $SU(2,q^2)$.
In analogy with the definition of $\mathfrak{su}(2)$, let us define $\mathfrak{su}(2,q^2)$ to be the space of all traceless $2\times 2$ matrices $x$ with entries in $F_{q^2}$, such that $\sigma(x)^T = -x$, where $\sigma$ is applied to every entry in $x$. Unless I am mistaken, $\mathfrak{su}(2,q^2)$ is $3$-dimensional over $F_q$.
Now the group $SU(2,q^2)$ acts on $\mathfrak{su}(2,q^2)$ by conjugation (the so-called Adjoint action). Let us check the non-trivial part of this claim:
$\sigma(gxg^{-1})^T = g \sigma(x)^T g^{-1} = -gxg^{-1}$.
Given an element $0 \neq x \in \mathfrak{su}(2,q^2)$, we define the ray passing through it as the set of all $\lambda x$, as $\lambda$ runs over all non-zero elements of $F_q$.
My question is now this: does the group $SU(2,q^2)$ act transitively on the set of all rays in $\mathfrak{su}(2,q^2)$? (I hope someone survived the length of this post)
Let $K$ denote the integers modulo $5$, and $L = K[x]/(x^2-2)$. $L$ is a quadratic extension of $K$, and is a field with $5^2$ elements. Denote by $\bar{x}$ the equivalence class of $x$ in $L$. Consider now the following matrix
$A = \left( \begin{array}{cc} \bar{x} & \bar{x} \\ \bar{x} &-\bar{x} \end{array} \right)$
Assume that $A$ was equivalent with respect to the Adjoint action, to a diagonal matrix $\Lambda$ in $\mathfrak{su}(2, 5^2)$ via an element $g \in SU(2, 5^2)$. In other words, we have:
$\Lambda = gAg^{-1}$.
Let us say that $\Lambda = \operatorname{diag}(\lambda \bar{x}, -\lambda \bar{x})$, for some $\lambda \in K$. But the matrices $A$ and $\Lambda$ have the same determinant, so that we must have
$-2\lambda^2 \equiv -4 \text{ mod 5 }$
which is equivalent to $\lambda^2 \equiv 2 \text{ mod 5}$, which in turn has no solution $\lambda \in K$. This answers negatively my question, so that the Adjoint action of a finite unitary group $SU(2,q^2)$ on the set of rays in its Lie algebra is in general not transitive.