What are the fixed-point free subgroups of $\mathrm U(2)$? Let $\mathrm U(2)\subset\Bbb C^{2\times 2}$ be the group of unitary $(2\times 2)$-matrices.
I wonder the following:

Question: What are the maximal fixed-point free subgroups of $\mathrm U(2)$?

A group $\Gamma\subset\mathrm U(2)$ is fixed-point free if there are no non-trivial $T\in\Gamma$ and non-zero $v\in\Bbb C^2$ with $Tv=v$. For example, $\mathrm U(2)$ itself is not fixed-point free, as
$$T:=\begin{pmatrix} -1 & 0 \\ \phantom+0 & 1 \end{pmatrix}\in \mathrm U(2)\setminus\{\mathrm{Id}\}$$
has the fixed point $(0,1)\in\Bbb C^2$.
I know that $\mathrm{SU}(2)\subset\mathrm U(2)$ is fixed point free (it acts like the multiplicative group of unit quaternions on $\Bbb H\cong\Bbb C^2$).
It is also maximal, since it acts regularly on the unit sphere $\Bbb S^1(\Bbb C)\subset\Bbb C^2$. 
There are also the conjugates of $\mathrm{SU}(2)$.
But are there any others?

Update
Thanks to the enlightening answer by YCor, I understand that the problem in this form seems to be intractable. By that answer I also realized that what I am more interested in are those maximal fixed-point free subgroups that have "interesting" finite subgroups, that is, other than, say, $\{\mathrm{Id}\}$ and $\{\pm\mathrm{Id}\}$. Or also, closed subgroups (in the topological sense).
YCor's answer already established that besides $\mathrm{SU}(2)$ we have the maximal torus $\mathrm U(1)\times\mathrm U(1)$ as another maximal subgroup.
Both are closed and have "interesting" finite subgroups.
 A: I can answer the question in the text "are there any others", the answer is yes, and I expect many, although I don't claim to answer the question in the title "What are the fixed-point free..." since it's not a full description.
Restated in more standard terms, the question is to understand the class $\mathcal{L}$ of subgroups of $\mathrm{U}(2)$ acting freely on $\mathbf{C}^2$ and especially describe the set $\mathcal{L}_\max$ maximal elements in $\mathcal{L}$. 
The OP already observed that $\mathrm{SU}(2)\in\mathcal{L}_\max$.
If $X$ is the closed subset of $\mathrm{U}(2)$ consisting of elements with $1$ as eigenvalue, observe that $\mathcal{L}$ consists of subgroups $G$ such that $G\cap X=\{\mathrm{id}\}$. 
Clearly every element of $\mathcal{L}$ is contained in a maximal one.
There are obvious cyclic subgroups in $\mathcal{L}$, namely the subgroup generated by any diagonal matrix $(u,v)$ with $|u|,|v|=1$ and $u,v$ not roots of unity. Choosing $uv\neq 1$, and embedding into a maximal subgroup, we obtain elements of $\mathcal{L}_\max$ that are not included in $\mathrm{SU}(2)$.
I also claim that any generic (in the topological or measure-theoretic meaning) pair freely generates a subgroup in $\mathcal{L}$. Indeed, for every nontrivial word $w(x,y)$, the $Y_{w}$ of pairs $(g,h)$ such that $w(g,h)\in X$ is Zariski-closed and is not the whole set of pairs (by evaluation to a free pair in $\mathrm{SU}(2)$). Hence the union over $w$ of $Y_w$ has measure zero, and is an $F_\sigma$ subset. For this we see that there are dense subgroups in $\mathcal{L}_\max$, and pushing a little it can be checked that it has dense subgroups of cardinal $2^{\aleph_0}$.
I guess there is no reasonable way to list elements of $\mathcal{L}_\max$. For instance, I expect that the cardinal of $\mathcal{L}_\max$ is the largest possible, namely $2^{2^{\aleph_0}}$; still I can check that this is the cardinal of $\mathcal{L}$. (But at the moment I don't even have a proof that $\mathcal{L}_\max$ modulo conjugation has cardinal $\ge 2^{\aleph_0}$.)
