Consider the General linear group $GL(n,q)$ over the finite field $\mathbb{F}_q$ of $q$ elements. The unitary group $U(n,q)$ is described as a subgroup of $GL(n,q^2)$ which is the set of linear maps which are invariant under a $c$-Hermitian form on the $n-$dimension vector space $V$ over $\mathbb{F}_{q^2}$, where $c:\mathbb{F}_{q^2}\to \mathbb{F}_{q^2}$ is the Frobenius automorphism $\alpha\mapsto \alpha^{q}$. Since, all the $c-$ Hermitian forms over $\mathbb{F}_q$ are equivalent, there is a unique isometry group upto isomorphism, which is $U(n,q)$.
Now, G.E Wall had described the conjugacy classes of these groups(over any field though). Now, I need to verify my understanding of conjugacy class of $U(n,q)$. This is what I understand:
(1) One defines a certain set of monic polynomials in $\mathbb{F}_{q^2}[x]$ as follows: If $f\in \mathbb{F}_{q^2}[x]$, then define the dual of $f$ as $\tilde{f}(x)=\frac{1}{f(0)^c}x^{deg(f)}f^{c}(x^{-1})$. In other words, $\tilde{f}$ is such that: $\alpha$ is a root of $f$ iff $\alpha^{-q}$ is a root of $\tilde{f}$. We call $f$ self-dual if $f=\tilde{f}$.
(2) Now, $K=\bigcup\limits_{g\in GL(n,q^2)} gU(n,q^2)g^{-1}$ is a normal subset of $GL(n,q^2)$, and hence a union of conjugacy classes of $GL(n,q^2)$. Now, by the theory of rational canonical forms, one can associate a combinatorial data to an element $T\in GL(n,q^2)$, as follows: If $\Phi$ is the set of all irreducible polynomials in $\mathbb{F}_{q^2}$ except $x$, then the combinatorial data of $T$ denoted by $\Delta_T$ is: for $\phi \in \Phi$, (a) a partition $\lambda_{\phi}$ of $|\lambda_{\phi}|$ (b) $\sum\limits_{\phi \in \Phi} deg(\phi)|\lambda_{\phi}|=n$. Now, By the theory of Wall, a conjugacy class $C$ of $GL(n,q^2)$ belongs to $K$ iff the combinatorial data is as follows: for $\phi \in \Phi$, (a) a partition $\lambda_{\phi}$ of $|\lambda_{\phi}|$ (b)$\lambda_{\phi}=\lambda_{\tilde{\phi}}$ (c)$\sum\limits_{\phi \in \Phi} deg(\phi)|\lambda_{\phi}|=n$. It is clear that $\phi$ is irreducible iff $\tilde{\phi}$ is irreducible.
(3) Finally, it also asserts the following: Suppose $C$ be a conjugacy class of $GL(n,q^2)$ residing in $K$. Then $C$ is as described in (2). Suppose $\emptyset \neq L\subset C$ is the part of $C$ contained in $U(n,q)$, then $L$ is a complete conjugacy class in $U(n,q)$, that is, it doesn't split further as $U(n,q)$-conjugacy class. It is implicit that for each conjugacy class $C$ as described above, there always exists a non-empty $L$ as above.
This completely describes the conjugacy class in $U(n,q)$. This kind of description can be found in several papers. For Example: In several papers of J.Fulman and others. But since the writing is much compact, I had expanded it in my own way. It is a big post and I will be very thankful if anyone can patiently read this and clarify me if my understanding is correct. Thank you!
