First, we will prove that 7 is possible. Consider the following hexagon $A B C D E F$ whose vertices are located at $A(0,0), B\left(\frac{1}{2}, 1-\frac{\sqrt{3}}{2}\right), C\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), D(0,1), E\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), F\left(-\frac{1}{2}, 1-\frac{\sqrt{3}}{2}\right)$. One can easily verify that all diagonals but $B E$ and $C F$ have length 1. Now suppose that there are at least 8 diagonals in a certain convex hexagon $A B C D E F$ whose lengths are equal. There must be a diagonal such that, with this diagonal taken out, the other 8 have equal length. There are two cases. Case I. The diagonal is one of $A C, B D, C E, D F, E A, F B$. WLOG, assume it is $A C$. We have $E C= E B=F B=F C$. Thus, $B$ and $C$ are both on the perpendicular bisector of $E F$. Since $A B C D E F$ is convex, both $B$ and $C$ must be on the same side of line $E F$, but this is impossible as one of $B$ or $C$, must be contained in triangle $C E F$. Contradiction. Case II: The diagonal is one of $A D, B E, C F$. WLOG, assume it is $A D$. Again, we have $E C=E B= F B=F C$. By the above reasoning, this is a contradiction. Thus, 7 is the maximum number of possible diagonals.