Finding a pair of exponents so that $c^a \bmod N = c^b \bmod N$ I am not very familiar with RSA, so this question may appear to be stupid. 

Suppose that in RSA encryption, we are given $c$ (the ciphertext) and $N$ (the semi-prime modulus), can we find a pair of exponents such that $c^a \bmod N = c^b \bmod N$?

This has nothing to do with decrypting RSA; I’m just curious whether it is possible to find a pair of exponents that can produce this result? Would this be feasible if the exponents are really large numbers (for example, 300 decimal digits or more)? If so, is there any computer algorithm that I can use to do this?
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.