Number of DFAs of only one state
First let $\Sigma = \{a,b\}$.

Assuming that you're taking a strict approach, there has to be a transition from every state for every symbol in the alphabet, then you have only two possible DFAs (that accept $\emptyset$ and $\Sigma^{\ast}$ respectively).

If you allow a partial transition function, where missing transitions imply rejection (note that this can still be interpreted deterministically), then there are five different languages that can be accepted (strictly there's seven automata): $\emptyset$, $\{\varepsilon\}$, $a^{\ast}$, $b^{\ast}$ and $\Sigma^{\ast}$.

With a larger alphabet, in the partial function case, you get more of course. With $|\Sigma| = n$ there are $2^{n} + 1$ possible languages, from $2^{n+1}$ different automata (half produce the same language). You get $2^{n}$ from each possible choice of which symbols will have a transition in the case where the state is accepting, and the $+ 1$ from the case where the state is non-accepting (it doesn't matter what transitions you have then, the languages are all $\emptyset$).