There can't be an equilibrium in which no one tells the police. In that case, it is a profitable deviation to tell since $v-c>0$.
There can't be an equilibrium in which more than one player tells the police. In that case, it is a profitable deviation for one of them to remain silent since $v>v-c$.
There are five pure strategy Nash equilibria in which only one player talks. This one player would be worse off by remaining silent, $0<v-c$, and the silent players would be worse off by telling, $v-c<v$.
Next, you consider mixed strategies. For the same reason as above, there is no equilibrium in which one or more players alway talk and others mix between talking and silence as pure silence is better given someone talks to the police for sure. Similarly, there is no equilibrium in which only one player randomizes and all others remain silent, as this one player would not be indifferent between both actions, talking is always better given the others don't.
There are equilibria in which two players randomize and the others are silent. Wlog, suppose player 1 and 2 talk with probabilities $p$ and $q$, respectively. For each player, talking always gives payoff $v-c$. Both players must be indifferent (otherwise they would not want to randomize). Hence,
$v-c = p v + (1-p)0 = q v + (1-q)0$ which we can solve for $p=q=\frac{v-c}{v}$. Given this indifference condition, it must also be a best response for player 3,4,5 to remain silent. This is true as $v-c\leq p^2 v + 2p(1-p)v$, because $v-c=pv<(p^2+2p(1-p))v$ and $p<(p^2+2p(1-p))$ for any $p\in(0,1)$. There are $\binom{5}{2}=10$ such equilibria.
Similarly, there should be $\binom{5}{3}=10$ and $\binom{5}{4}=5$ equilibria in which 3 and 4 players talk and the others are silent, and one equilibrium in which all five players randomize over both options with some probability $z$ solving $v-c=(1-(1-z)^4)v$.
All in all, there should be $5+10+10+5+1=31$ Nash equilibia.
