This is a very nice exercise in using Clifford theory. For our purposes, we will need the following two facts:
- If $N$ is normal in $G$ with $G/N$ cyclic, then an irreducible character $\chi$ of $N$ arises as the restriction of an (irreducible) character of $G$ if and only if $\chi$ is fixed by the action of $G/N$ on the set of irreps of $N$. This action is given by $$\bar{a}\cdot \chi(g):=\chi(ag a^{-1})$$
- If $\chi$ is an irrep of $N$, not fixed by the cyclic group $G/N$, then $Ind_N^G(\chi)$ is an irrep of $G$.
So now both the generalised Quaternion group and the Dihedral group have a normal cyclic subgroup $N$ of order $2^{n-1}$, and any element of $G$ not in this subgroup acts as inversion $a\mapsto a^{-1}$ when conjugating elements in $N$. The only difference between these two groups is that in one case, we can find some external (not in $N$) $g$ that squares to $e$ (dihedral), or we can at best find an external $g$ that squares to the unique element of order $2$ in $N$.
Either way, the action of this quotient on the irreps of $N$ will be the $C_2$ action which maps each irrep to its dual, since this is what inversion $a\mapsto a^{-1}$ does on characters. So "over $N$", these two groups look the same, their conjugacy classes are orbits of inversion on $N$, and their two dimensional irreps correspond under this bijection, since these have support only on the classes contained in $N$, and on these classes, they are the sums $\lambda+\bar{\lambda}$, for $\lambda$ not self dual.
So now we need to check the linear characters, which means checking the abelianisation of these two groups. Since we know there are four linear characters (two lying over each of the two linear characters of $N$ fixed by our inversion), we know there will be exactly two conjugacy classes in $G- N$, and our abelianisation will be isomorphic to the elementary abelian group of order $4$. The characters of these are just $\pm 1$, and one can check that a bijection of these external classes will agree on these classes, this is simple to check by hand, its exactly the same as the order $8$ case. So their character tables agree, and we are happy.
To summarise, we can use the large normal abelian subgroup $N$, and check how irreps of $G$ restrict to $N$. Using Clifford theory, we see at once from the usual presentations of these groups that the nonlinear characters of these groups are naturally in bijection with size $2$ orbits of the inversion map on the irreps of $N$, so we get that most of the character tables agree, and then we just check that the remaining four linear characters agree, which isn't too tricky since they are all order $2$. If this last part isn't clear, I can expand on it, but once you know the strategy, the proof of all of this isn't too hard, and is a good exercise to do.
Best Answer
The two nonisomorphic nonabelian groups of order $p^3$ for odd primes $p$ have the same character tables and the same Frobenius-Schur indicators - these are $0$ for all except the trivial character. I would guess that the nonabelian groups of order $27$ are the smallest such examples, but I haven't checked.