Expected length of orbit for random permutation

The probability that $\sigma(1)=1$ (so $m=1$) is $\frac{1}{n}$. Conditional on $\sigma(1)\neq1$, we know that $m=2$ with probability $\frac{1}{n-1}$, so the unconditional probability is $\frac{n-1}{n}\cdot\frac{1}{n-1}=\frac{1}{n}$. Continuing like this, we see that $\mathbb P(m=k)=\frac{1}{n}$ for any $1\leq k\leq n$. So $$\mathbb E[m]=\frac{1}{n}(1+2+\dots+n)=\frac{n+1}{2}.$$

Choose the elements in the cycle of $1$ say $m-1$ order them and then the rest of the elements are other permutation so there are $(m-1)!(n-m)!$ of them. Then $$\mathbb{E}[m]=\sum _{m=1}^nm\cdot P(\sigma \text{ has $1$ in a cycle of length m})=\sum _{m=1}^nm\cdot \frac{\binom{n-1}{m-1}\cdot (n-m)!(m-1)!}{n!}=\sum _{m=1}^n\frac{m}{n}.$$ Your conjecture is obtained by the Hockey-Stick identity.