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](https://mathscinet.ams.org/mathscinet-getitem?mr=2680716) 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)



Finite groups with the same character table *including* class types, and square-free order
(Turning my comments into an answer). 


A finite group with all Sylow subgroups cyclic is called a *$Z$-group*. 


According to review MR0470050 in MathSciNet, in [1] it is shown that a $Z$-group is determined by its character table. So the answer to question 2 would be no.


I do not have access to [1] and I have not seen it, if you are interested it seems you might have to look for a physical copy in a library somewhere. 


However, perhaps with some effort you could figure out your own proof. We have a good understanding of the structure of $Z$-groups (in particular of groups of squarefree order). See for example [2] for an enumeration of all $Z$-groups of given order. I would guess the character tables of $Z$-groups are also known.


[1] Çelik, Özdem. On the character table of Z-groups. Mitt. Math. Sem. Giessen Heft 121 (1976), 75–77


[2] M.R. Murty and V.K. Murty. On groups of squarefree order. Math. Ann. 267 (1984), 299–309.


