Solution:
The answer is four.

First we show that three candies are not always enough. If you only buy three candies, then it is possible the three coins you spend will be some combination of $1$-, $2$-, and $5$-florin coins, in which case you definitely won't receive any $10$- or $20$-florin coins in change. Thus, in this situation, you do not get any information that can be used to distinguish the $10$- and $20$-florin coins.

Now we show that four candies are enough. In each of these transactions, pay with a different kind of coin. If any of the transactions does not produce change, then you must have paid in that transaction with a $1$-florin coin. If all the transactions produce change, then the coin you didn't use as payment is the $1$-florin coin. In either case, you can identify the $1$-florin coin. Since the change for $2$ florins will be $1$ florin, you can now tell if you paid for any of the transactions with a $2$-florin coin, and thereby identify which of the five coins has a value of $2$ florins. Next, we know that the change for $5$ florins will be either two $2$-florin coins or four $1$-florin coins, or one $2$-florin and two $1$-florin coins; in any event, you can recognize these collections of coins and then deduce which coin is the $5$-florin coin (if you see one of these collections, then you know which coin was the $5$-florin coin; otherwise, you know it was the coin you didn't use as payment). Likewise, change for $10$ florins will be a collection of coins that total $9$ florins, and all such collections can now be recognized. Thus we can deduce which coin is the $10$-florin coin, and finally we can apply the same procedure to deduce the identity of the $20$-florin coin.