The expected number of fixed points of a random permutation

1. For each $i$, $\sigma(i)$ is uniformly distributed on $\{1,\dots,n\}$, so $\mathbb{E}[I_i]=\Pr[\sigma(i)=i]=\frac1n$. By linearity,
$$\mathbb{E}[X]=\sum_{i=1}^n\frac1n=1.$$

2. $X(X-1)=\sum_{i\ne j}I_iI_j$, the number of ordered pairs $(i,j)$, $i\ne j$, of fixed points. For $i\ne j$,
$$\Pr[\sigma(i)=i,\ \sigma(j)=j]=\frac{(n-2)!}{n!}=\frac1{n(n-1)}.$$
There are $n(n-1)$ ordered pairs, so $\mathbb{E}[X(X-1)]=n(n-1)\cdot\frac1{n(n-1)}=1$.

Variance. $\mathbb{E}[X^2]=\mathbb{E}[X(X-1)]+\mathbb{E}[X]=2$, so $\operatorname{Var}X=2-1^2=1$.

Check. For $n=5$ a computer enumeration of the $120$ permutations gives exactly $\mathbb{E}[X(X-1)]=1$.

Remark. The fixed points are far from independent (if $n-1$ points are fixed, so is the last one), yet linearity of expectation does not care. For $n=2$: $X$ is $2$ or $0$ with probability $\frac12$ each.

Lesson. Write the quantity as a sum of indicators and use linearity of expectation: no independence needed.