Mean number of fixed points of random function from $\{1,2,3,\ldots,n\}$ to $\{1,2,3,\ldots,n\}$ Expected value (mean) tends to confuse me lately.

Let $ \Omega$ be the discrete space of all functions $ \omega:\{1,2,3,\dots,n\} \rightarrow\{1,2,3,\dots,n\}$, $\mbox{Pr}$ is uniformly distributed.
  Let $f$ be a random variable that counts all fixed points of all functions in $ \Omega $
  $$ f(\omega) = |\{i \in \{1,2,3,...,n\}: \omega(i)=i\}$$

Well, I had two ideas which ended the same way.
In both of them I defined n indicators $f_1,f_2,f_3...f_n$, such that for each $ 1\leq i\leq n$, return 1 if $f(i)=i$ and 0 otherwise.
Than, the expected value is the sum of $ \sum_{i=1}^{n} f_i$
First idea
Looking at index i, the probability to choose i as an image is $Pr(fi=1)= \frac{1}{n}$ then the expected value is $\frac{1}{n}+\frac{1}{n}+\frac{1}{n}...+\frac{1}{n}=\frac{n}{n}=1$
Second idea
For index i there is one option, and for the rest of the indexes are free to be chosen, so $Pr(fi=1) = \frac{1 \cdot n^{n-1}}{n^n}$
Then the expected value is once again the sum, which is $ n \cdot \frac{n^{n-1}}{n^n} = \frac{n}{n}=1 $
Am I on the right track or totally lost?
 A: By way  of enrichment  let me point  out that  this can be  done using
simple  combinatorial   species  as   shown  at  the   following  MSE
link.
Start from the species of labelled trees has the specification
$$\mathcal{T} = 
\mathcal{Z} \times \mathfrak{P}(\mathcal{T})$$
which gives the functional equation
$$T(z) = z \exp T(z).$$
We have that $$T(z) = \sum_{n\ge 1} n^{n-1} \frac{z^n}{n!}.$$
To  compute the  expected number  of fixed  points note  that these
random  mappings are  sets  of cycles  of  trees having  combinatorial
specification
$$\mathfrak{P}
\left(\sum_{q\ge 1} 
\mathfrak{C}_{=q}(\mathcal{T}(\mathcal{Z}))\right).$$
Now observe that a fixed point is in fact the root of a tree placed in
a one-cycle. This gives the marked species
$$\mathfrak{P}
\left(\mathcal{U} \mathfrak{C}_{=1}(\mathcal{T}(\mathcal{Z}))
+ \mathfrak{C}_{\ge 2}(\mathcal{T}(\mathcal{Z}))\right).$$
We get the generating function
$$G(z, u) =
\exp\left(u T(z) + \sum_{q\ge 2} \frac{T(z)^q}{q}\right)
\\ = \exp\left(u T(z) - T(z) + \sum_{q\ge 1} \frac{T(z)^q}{q}\right)
\\ = \exp\left(u T(z) - T(z) + \log\frac{1}{1-T(z)}\right)
\\ = \frac{1}{1-T(z)} \exp(uT(z)-T(z)).$$
To get the expectation compute
$$\left. \frac{\partial}{\partial u} G(z, u)\right|_{u=1}$$
which yields
$$\left. \frac{1}{1-T(z)} \exp(uT(z)-T(z)) T(z)
\right|_{u=1}
= \frac{T(z)}{1-T(z)}.$$
The desired value is then given by
$$n^{-n} \times n! \times [z^n] \frac{T(z)}{1-T(z)}
= \frac{n^{-n}\times n!}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{T(z)}{1-T(z)} \; dz.$$
Putting $w=T(z)$ we get from the functional equation $w=z\exp(w)$
or $z=w\exp(-w)$ and $dz = (\exp(-w)-w\exp(-w)) dw$ which yields
$$\frac{n^{-n}\times n!}{2\pi i} 
\int_{|w|=\gamma} \frac{\exp(w(n+1))}{w^{n+1}} 
\frac{w}{1-w} \exp(-w)(1-w) \; dw
\\ = \frac{n^{-n}\times n!}{2\pi i} 
\int_{|w|=\gamma} \frac{\exp(wn)}{w^{n}}  \; dw
\\ = n^{-n} \times n! \times \frac{n^{n-1}}{(n-1)!}
= n^{-1} \times n = 1.$$
The purpose  here was to  introduce the species  which can be  used to
extract a variety of additional non-trivial statistics.
