If $e^{itA}$ is a special unitary matrix, what can I say about A? I know that if $e^{itA}$ is unitary, A must be skew-hermitian. I also know that $A$ must be traceless, since the determinant of $e^{itA}$ is $1$. 
What can I say about the determinant of A?
 A: [Begin EDIT]
The following answer is correct under the additional assumption that the matrix $A$ is "nice" enough. As such, it describes a set of matrices with this property, but not all of them. I leave the answer unedited below, and add a discussion of a counterexample below.
[End EDIT]
I am assuming (from context clues) that $t\in\mathbb R$. Now
$e^{itA}$ is unitary if and only if
$$
(e^{itA})^\dagger e^{itA}=I.
$$
It is easy to see (for instance via its Taylor series) that $(e^{itA})^\dagger=e^{-itA^\dagger}$.
Thus, the unitarity condition is that
$$
e^{it(A-A^\dagger)}=I.
$$
Thus $e^{itA}$ is unitary whenever $t(A-A^\dagger)$ is diagonalizable with all eigenvalues in $2\pi\mathbb Z$. In particular, if $A$ is Hermitian ($A=A^\dagger$) this condition holds, but that is not the only way for $e^{itA}$ to be unitary.
The determinant of $A$ plays no role in this condition. (The trace is used to enforce special unitarity, as you already observed correctly.)
[Begin EDIT]
As pointed out in the answer by user1551, there are matrices satisfying the condition for which $e^{-itA^\dagger}e^{itA}\not=e^{it(A-A^\dagger)}$, which my argument does not cover. My argument covers the case when $A$ is a normal matrix (that is, $A$ and $A^\dagger$ commute) but this need not be true in general.
In fact, I was quite surprised to see the counterexample posted by user1551, and it was not immediately obvious to me that it was in fact a counterexample. In case it helps other readers, here is the explanation why.
Fix two integers $m$ and $n$ with $|m|\ne|n|$, and let
$$
D=\pmatrix{2m\pi&0\\ 0&2n\pi},\qquad N=\pmatrix{0&1\\ 0&0},\qquad A=D+N.
$$
Observe that $D^aND^bND^c=0$ for any $a,b,c$, since $DN$ and $ND$ are both scalar multiples of $N$ and $N^2=0$. Thus, by the non-commutative version of the binomial theorem,
$$
A^k=D^k+\sum_{i=1}^k D^{i-1}ND^{k-i},\qquad k\geq 1.
$$
Now observe that $D^{i-1}ND^{k-i}=(2\pi)^{k-1}m^{i-1}n^{k-i}N$. Thus,
$$
A^k=D^k+(2\pi)^{k-1}\sum_{i=1}^k m^{i-1}n^{k-i}N,
$$
whereupon from the Taylor series for the matrix exponential we have that
$$
e^{iA}=e^{iD}+\sum_{k=1}^{\infty}\frac{i^k(2\pi)^{k-1}}{k!}\sum_{i=1}^km^{i-1}n^{k-i}N.
$$
$$
=I+\sum_{k=1}^{\infty}\frac{i^k(2\pi)^{k-1}}{k!}\frac{m^k-n^k}{m-n}N
$$
$$
=I+\frac{e^{2\pi m i}-e^{2\pi n i}}{2\pi(m-n)}N=I.
$$
[End EDIT]
A: Presumably $t$ is a nonzero real number. Since matrix exponentials of non-trivial Jordan blocks are not diagonalisable over $\mathbb C$, if $e^{itA}$ is unitary, $A$ must be diagonalisable and every eigenvalue $\lambda$ of $A$ satisfies $|e^{it\lambda}|=1$. Hence each $\lambda$ is real, i.e. $A$ is a diagonalisable matrix with real eigenvalues, and $|\det(e^{itA})|=1$ because $e^{itA}$ is unitary.
However, we cannot infer that $A$ is traceless or Hermitian. E.g. for any two integers $m$ and $n$ with $|m|\ne|n|$, the matrix
$$
A=\pmatrix{2m\pi&1\\ 0&2n\pi}
$$
is neither traceless nor symmetric, but $e^{iA}=I$ is real orthogonal. The determinant of $A$ is also unbounded in this example.
A: In fact, knowing the unitary matrix $e^{iA}$, we can explicitly calculate the associated matrices $A$. 
Clearly, we may put $t=1$ and we may assume that $spectrum(A)=(\lambda_j)_j$ and (up to an orthonormal change of basis) $e^{iA}=diag(e^{i\mu_1}I_{k_1},\cdots,e^{i\mu_r}I_{k_r})$, where the $e^{i\mu_j}$ are the distinct values of the $e^{i\lambda_j}$. $|e^{i\lambda_j}|=1$ implies that $\lambda_j\in\mathbb{R}$ (cf. user1551's post).
Since $A,e^{iA}$ commute, $A=diag(A_1,\cdots,A_r)$ where $e^{iA_j}=e^{i\mu_j}I_{k_j}$, that is, $A_j=P_jdiag(\nu_1,\cdots,\nu_{k_j}){P_j}^{-1}$ with $e^{i\nu_1}=e^{i\mu_j}$ and $\nu_l-\nu_1\in 2\pi\mathbb{Z}$ (note that $P_j$ is not necessarily unitary). 
Remark that, GENERICALLY (randomly choose $A$), the eigenvalues of $e^{iA}$ are distinct, and, consequently, $A$ is also diagonal; in other words, $A$ is diagonalizable in an orthonormal basis, and then, is normal.
