I just now noticed this question. Of course dhy is correct, as usual. I want to remark that there are also examples that arise "in nature". Let $X$ be any smooth, projective hypersurface in $\mathbb{P}^n$ of degree $d$. If $3\leq d \leq n$, then there exist lines $L\subset X$ such that $N_{L/X}$ is not nef. If the characteristic is $0$, or even just $>d$, and if $d\leq n-1$, then for a general line $L'$ in $X$, $N_{L'/X}$ is nef, i.e., $L'$ is "free". Thus some lines in $X$ have nef normal sheaf, while others do not.

If $X$ is sufficiently general, then the Fano subscheme $F\subset \text{Grass}(\mathbb{P}^1,\mathbb{P}^n)$ parameterizing lines contained in $X$ is smooth of dimension $2n-d-3$ with dualizing sheaf $\omega_{F/k} \cong \mathcal{O}_{G}(-n-1+d(d+1)/2)|_F$, where $\mathcal{O}_G(1)$ is the ample generator of the Picard group of $\text{Grass}(\mathbb{P}^1,\mathbb{P}^n)$. With the single exception of $(n,d) = (3,2)$, if $2n-d-3$ is positive then $F$ is connected. In particular, if $d(d+1)/2 \leq n$, with the two exceptions $(n,d) = (2,1)$ and $(3,2)$, then $F$ is a Fano manifold. Thus, by Campana and Kollár-Miyaoka-Mori (or just direct computation in this explicit case), $F$ is rationally chain connected. Therefore any two lines in $X$ are rationally equivalent, yet some have nef normal sheaf and others do not.