Problem related to the discrete logarithm problem Let $G$ be a generator of a cyclic group in which the discrete logarithm problem is hard and $x$ and $u$ be scalars of the group such that $X = xG$ and $U = uG$, respectively. We want to compute $J = x^{-1}U$. Is it possible to calculate it without knowing $x$, even if we know $u$?
This problem is equivalent to the Computational Diffie Hellman problem; this remains true even if we know $u$.
See this paper for details; in summary:

Suppose we have an oracle that, given $G$, $X = xG$ and either (depending on the oracle type) $U = uG$ or $u$, gives us the value $J = x^{-1}uG$.  Then, we can use this Oracle to construct a second Oracle that, given $H, aH$, computes $a^2H$ (by passing $G = aH$, $X = H = a^{-1}G$, $U=G$ (or $u=1$), and recovering $J = (a^{-1})^{-1} \cdot 1 \cdot aH = a^2H$.  We can then use this second Oracle to solve the CDH problem (given $G, aG, bG$, we compute $(2ab)G = (a+b)^2G - a^2G - b^G$, and then do a point halving.

Suppose we have an oracle that solves the CDH problem, that is, given $H, aH, bH$, gives us the value $abH$.  Then, given $G$, $xG$, we can compute $x^{-1}G$ by setting $H = xG$, $aH = bH = G = x^{-1}H$ and $bH = H$.  We give this to the Oracle which gives us the value $(x^{-1} \cdot x^{-1}) H = x^{-1}G$.  Then, we can convert that value into $J$ by either multiplying it by $u$ (if we know it), or invoking our CDH Oracle one more time with the inputs $G, x^{-1}G, uG$.