Is a finite group with any subgroup admitting a unique complement, cyclic? Let $G$ be a finite group such that any subgroup $H \le G$ admits a unique complement $K$.  
Question: Is $G$ cyclic ?
 A: This result admits a proof at the subgroups lattice level. 
Let $\mathcal{L}(G)$ be the subgroups lattice of $G$, then we see that $\mathcal{L}(G)$ does not admit sublattice equivalent to the following diamond or pentagon lattices

But there is the following result (see theorem 101 p 109 in this book): 

Theorem: A lattice is distributive iff it has no sublattice equivalent to the diamond lattice or the pentagon lattice.

It follows that the lattice $\mathcal{L}(G)$ is distributive.  
The result follows by a result of Oystein Ore of 1938 (theorem 4 p 267 in this paper): 

Theorem: A finite group is cyclic iff its subgroups lattice is distributive.

Remark: we observe that $\vert G \vert$ is square-free.
Remark: as a corollary of the proof, we see that the more generalized statement for which we replace "complement" by "lattice-complement" (i.e. $\langle H,K \rangle  =G $ and $H \cap K = 1$) is also true.
A: Take $p$ a prime dividing $|G|$ then a $p$-Sylow $S$ admits a unique complement $H$. The cardinal of $H$ is prime to $p$. Now take $S'$ another $p$-Sylow, it is clear that it will be a complement of $H$ as well (the intersection with $H$ is trivial and then $S'H$ cannot but be $G$ by cardinality). Hence $G$ admits a unique $p$-Sylow since they are all complement of the same $H$. 
From this it follows that :
$$G=S_{p_1}\times...\times S_{p_r} $$ 
$G$ is a product of its $p$-Sylows. Now because of this decomposition, the property holds for $G$ if and only if it holds for any of its $p$-Sylows.

Assume $S$ is a non-abelian $p$-group then its center does not admit any complement.

Assume $G=Z(G)H$ then $H$ is a non-trivial $p$-group, hence admits a non-trivial center which must be included in $Z(G)$ (any $G$ is $zh$ where $z$ is central and $h$ commutes with any element of the center of $h$) contradicting the complement assumption.

Assume $S$ is an abelian $p$-group non-cyclic then the group generated by an element of maximal order admits at least two different complements.

Use the decomposition of abelian groups.

Assume $S$ is a cyclic $p$-group then no non-trivial proper subgroups of $S$ admits a complement.

Use the unicity of subgroups of given order in cyclic groups.
Using those last three results we get that a $p$-group satisfying your condition is necessary cyclic of order $p$. 
In particular $G$ is cyclic.  
