Assume we have a semi-prime $N = pq.$ Assume $N$ is not divisible by $2$ or $5$. We want to solve the equation $$10^x \equiv 1 \mod N, \ \ \ x>0.$$ One solution is enough.
Is there any fast method for doing this without knowing $p$ and $q$?
A "trivial" solution would be $$x = N - (p-1)(q-1),$$ but this requires knowing $p$ and $q$.
One method would be to test every number up to $N - (p-1)(q-1)$ or maybe to start from $N-1$ and go down from there, it would require $p+q$ trials.
Is this as hard a problem as integer factorization?
The problem of finding the smallest $x > 0$ for this is finding the multiplicative order of $10 \bmod N$, and no efficient algorithm is known.
An $x$ here can be used to obtain the factorization of a number that isn't a prime power, and this is in fact how the end of Shor's Algorithm works, assuming $x$ isn't odd and $10^{x/2}$ isn't $-1$, in which case you need to try a different coprime base than 10. Assuming you had such an $x$, you could do $10^{x/2}$, which would lead to a number $a \not\equiv \pm 1 \pmod N$ such that $a^2 \equiv 1 \pmod N$. Then $(a + 1)(a - 1) \equiv 0 \pmod N$, and we can find the factors via GCD($a + 1, N$) and GCD($a - 1, N$).
For any $x > 0$, a multiple of the multiplicative order, we could just divide it by two until we get a number we can work with or can conclude that we need a different base (e.g. getting an odd number $x$ such that $10^x$ that still equals $1$ or that $10^x$ is equal to $-1$).
So, given the ability to solve this for 10 and other bases, you could factor, so at least that generalization isn't any easier.