Let $q=p^{f}$ be a prime power and $U_{n}(q):=\left\{A\in \mathrm{GL}_{n}(\mathbb{F}_{q^{2}}):A^{*}A=I\right\}$ where $A^{*}$ is the conjugate transposition of $A$(the conjugation action is the unique non-trivial Galois action of $\mathbb{F}_{q^{2}}/\mathbb{F}_{q}$).
I guess the abelianization of $U_{n}(q)$ is isomorphic to $C_{q+1}\cong \left\{a\in \mathbb{F}_{q^{2}}^{\times}:\overline{a}a=1\right\}$ where $\overline{a}$ is the conjugation of $a\in \mathbb{F}_{q^{2}}$. The reason of my guess is that the image of the homomorphism $\det:U_{n}(q)\to \mathbb{F}_{q^{2}}^{\times}$ from $U_{n}(q)$ to the abelian group $\mathbb{F}_{q^{2}}^{\times}$ is $\left\{a\in \mathbb{F}_{q^{2}}^{\times}:\overline{a}a=1\right\}$.
I wonder whether my guess is right and a rigorous proof. You may assume that $p$ is odd if you want.
You are right, but any proof is likely to require the simplicity of $\mathrm{PSU}_n(q)$ (for $n\geq 3$ and $(n,q)\neq (3,2)$). Given that, your statement, plus the fact that $\mathrm{SU}_n(q)$ is equal to its own derived subgroup, completes the proof.