Finding a pair of exponents so that $c^a \bmod N = c^b \bmod N$
If you find such you can break RSA. 

Assuming the order of c is maximal(likely if c is random ciphertext).
If we have such distinct exponents  and  they must be equal mod $\lambda(n) = lcm(p-1,q-1)$.

Let the difference between the exponents be $f=b-a$, $f=0\space mod \lambda(n)$

We pick a value e coprime with e, pick a small prime and verify.
We will then calculate $d = e^{-1} mod f$ 

We can then find the factorization of N using the algorithm described: https://www.di-mgt.com.au/rsa_factorize_n.html 

So no, if you do not know the factorization of $n$ you can't find such an exponent pair.

Example: 

N= 25777 //from link
c= 1517 //randomly chosen
a = 111 //less randomly chosen(I picked)
Possible b are: 6475,12839,19203,25567 we will pick the last one
this gives us f=25456
We try e=3 and verify indeed gcd(e,f)=1
We use extended eucleadean algorithm to find d=16689 and this is exactly the value used in the above link to factorize n.

If we pick b=19203
we get f=19092 and e=3 doesn't work so we will pick e=5 
We get d=7637 

we follow the algorithm linked
k=de-1 = 38184

We try several g, skipping to g=5
t=k=38184
g^t % n = 1 
t=t/2 = 19092
g^t % n = 1 
t=t/2 = 9546
g^t % n = 15050
and we check gcd(15050,n) = 149 


We found a non trivial factor of n. Q.E.D

Edit 2: Fgrieu comments the proof above does not hold in the general case. For some N the first assumption of high order may not be sound, and for very large b-a the last algorithm may not be efficient. I do not know how to rectify these. Which means though this "proof" should still be sufficient evidence that the original problem in the question of finding such a pair a,b is indeed very hard. Yet this is not a solid reduction between that problem and factorization in the general case.