It is known that two finite groups with the same complex character table are not necessarily isomorphic. The examples of smallest order are the quaternion group and the dihedral group of order $8$.
We can extend the character table by specifying the power maps. Given an element $x\in G$, we can calculate any $x^n$ up to conjugacy by following an element through its power maps one prime factor at a time.
We can define this because two conjugate elements raised to the same power will still be conjugate: $$\begin{align}x^n &= h^{-1}(g^{-1}xg)^nh\\ &= h^{-1}g^{-1}(x(gg^{-1}))^{n-1}xgh^{-1}\\&= h^{-1}g^{-1}x^{n}gh\end{align}$$$$, \forall n\in\mathbb{N}, \forall x,g \in G,\exists h\in G $$
. Then the aforementioned groups of order $8$ have nonisomorphic power maps: $$\begin{bmatrix} Q_8 & 1A & 2A & 4A & 4B & 4C \\ 2 & 1A & 1A &2A &2A &2A \end{bmatrix}$$ $$\begin{bmatrix} D_{4} & 1A & 2A & 2B & 2C & 4A \\ 2 & 1A & 1A & 1A &1A & 2A \end{bmatrix}$$
I have checked every pair of groups with isomorphic character tables up to order $27$, and they have different power maps. I know that's not much, but I can't imagine stumbling into a counterexample and I'm sure an expert would have far more efficient tactics to find one.
The question:
What are two non-isomorphic finite groups $G_1, G_2$ with isomorphic character tables, including power maps?
