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Yes, it is equally as difficult; if we assign: $$g' = ag$$ $$a' = a^{-1}$$ $$b' = b$$ Then the restatement of your problem is: given $g' = ag$, $a'g' = g$ and $b'g' = abg$, compute $a'b'g' = bg$, which is exactly the ECDH problem. Now, this assumes that $a$ has an inverse; this is not a problem if the curve order is a prime, and is easy to work around if ...