There are non-isomorphic finite groups with the same (complex) character table, as $D_4$ and $Q_8$.

$$\scriptsize\begin{array}{c|c} \text{class}&1&2A&2B&2C&4 \newline \text{size}&1&1&2&2&2 \newline \hline \rho_1 &1&1&1&1&1 \newline \rho_2 &1&1&-1&1&-1 \newline \rho_3 &1&1&1&-1&-1 \newline \rho_4 &1&1&-1&-1&1 \newline \rho_5 &2&-2&0&0&0 \newline \end{array} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{c|c} \text{class}&1&2&4A&4B&4C \newline \text{size}&1&1&2&2&2 \newline \hline \rho_1 &1&1&1&1&1 \newline \rho_2 &1&1&-1&1&-1 \newline \rho_3 &1&1&1&-1&-1 \newline \rho_4 &1&1&-1&-1&1 \newline \rho_5 &2&-2&0&0&0 \newline \end{array}$$ But the character tables of $D_4$ and $Q_8$ are no more equal if we include the class types, as $(1,2A,2B,2C,4) \neq (1,2,4A,4B,4)$. A class is of type $nX$ if its elements has order $n$.

**Question 1**: Are there non-isomorphic groups with the same character table *including* class types?

*Answer*: Yes (see the comment of Derek Holt) and if in addition their conjugacy classes have the same power map, they are called Brauer pairs. Among the $2$-groups, the smallest order of a group in a Brauer pair is $2^8$, and among the $56092$ groups of order $2^8$, there are exactly ten Brauer pairs (see MR2680716 Theorem 2.6.2 page 136).

I am specifically interested in groups of square-free order.

**Question 2**: Is there a Brauer pair of square-free order groups?

(unless any two square-free order groups with same character table are isomorphic)