Still expecting my fix

What’s the expected number of fixed points of a uniformly distributed random permutation of $n$ elements?

1. Consider a permutation $\sigma$ of $[n]$, and let $X_n^j$ be the random variable that takes the value $1$ if this permutation fixes $j$, and takes the value zero elsewhen. Set $X_n= \sum_{j=1}^n X_n^j$.
For every $j\in [n]$, $P(X_n^j=1)=(n-1)!/n!=1/n$. By the linearity of the spected value, $E[X_n]= \sum_{j=1}^nE[X_n^j] = \sum_{j=1}^n1/n =n/n=1$. So, the expected number of fixed points is $1$ independently of $n\in\mathbb{N}$.
Let’s compute its variance. If $n=1$, obviously $Var[X_1]=0$. In the case $n>1$ we have $Var[X_n]=E[(X_n)^2]-E[X_n]^2 = E[(\sum_{j=1}^n X_n^j)^2]-1 = E[\sum_{j=1}^n (X_n^j)^2] + 2E[\sum_{j\neq i}X_n^jX_n^i] -1$
But $(X_n^j)^2=X_n^j$ and $X_n^jX_n^i$ takes the value $1$ if $i$ and $j$ are fixed points and takes the value $0$ otherwise. Thus $Var[X_n] = E[\sum_{j=1}^n X_n^j] + 2\sum_{j\neq i}E[X_n^jX_n^i] -1 = 1 +2\sum_{j\neq i}(n-2)!/n! -1 = 2\sum_{j\neq i}1/(n-1)n = 2\binom{n}{2}1/(n-1)n = 2(n!)/2!(n-2)!(n-1)n = 1$
So, for $n>1$ its variance is also $1$ independently of $n$.