Question: If $n^r(X_n-X)$ converges in distribution for some $r>0$, do we have that $X_n$ converges to $X$ in probability?
As $P(n^r|X_n-X|>\varepsilon)\to P(|Z|>\varepsilon)$ (where $Z$ is the random variable to which $n^r(X_n-X)$ converges), can we say something about $P(|X_n-X|>\varepsilon)$?
(I take it that you assume that the random variables $X_n-X$ are defined on a common probability space; otherwise it doesn't make sense to ask about convergence in probability.)
Fix $\epsilon>0$. For any $\delta>0$ there exists $R=R(\delta)>0$ such that
$$\mathbb{P}(|Z| >R) \leq \delta; \tag{1}$$
since the distribution function of $|Z|$ has at most countably many discontinuity points, we may assume without loss of generality that $R$ is a continuity point. If we choose $N \in \mathbb{N}$ large enough such that
$$\epsilon N^r \geq R \tag{2}$$
then we have for all $n \geq N$
$$\begin{align*} \mathbb{P}(|X_n-X| >\epsilon) &= \mathbb{P}(n^r |X_n-X| > \epsilon n^r) \stackrel{(2)}{\leq} \mathbb{P}(|X_n-X| > R). \end{align*}$$
As $R$ is a continuity point we find from the weak concergence that
$$\limsup_{n \to \infty} \mathbb{P}(|X_n-X|>\epsilon) \leq \limsup_{n \to \infty} \mathbb{P}(|X_n-X| > R) = \mathbb{P}(|Z|>R) \stackrel{(1)}{\leq} \delta.$$
As $\delta>0$ and $\epsilon>0$ are arbitrary, this proves $X_n-X \to 0$ in probability.