Suppose $n$ is odd. There are two guests with the same number of acquaintances. If they do not have a common acquaintance or a common unknown, then each of them knows exactly $\frac{n-2}2$ other people, which is not an integer, contradiction. Hence Fredek is right for odd $n$. Suppose Fredek is wrong for some $n\ge6$ even. Take two people $A$ and $B$ with the same number of acquaintances. Each of them has either $\frac{n}2$ or $\frac{n-2}2$ acquaintances. Among guests other than $A$ and $B$, there exist $C$ and $D$ with the same number of acquaintances. Each of $C$ and $D$ knows exactly one of $A$ and $B$, thus they have either $\frac{n}2$ or $\frac{n-2}2$ acquaintances. Similarly, there exist $E$ and $F$ with either $\frac{n}2$ and $\frac{n-2}2$ acquaintances. Among $A,B,C,D,E,F$, there exist three with the same number of acquaintances. Among these three, one is a common acquaintance or a common unknown for the other two, so Fredek is right. For $n=4$, Fredek is wrong: A only knows B, B only knows A and C, C only knows B and D.