Is it possible to decrypt a ciphertext with a different private key? In public key cryptosystem, there are often two keys (pub and pri) and two functions (Enc and Dec) such that:

Enc(pub, m) = c
Dec(pri, c) = m


Usually pub and pri are generated as a matching key pair and used together. However, given a ciphertext c, without knowing pri, is it possible to successfully decrypt it into the original plaintext m with a different private key pri' other than pri?
No, it's not possible. If it was possible it would have a devastating impact on asymmetric cryptography in general.

Most asymmetric cryptosystems rely on mathematically problems that cannot be solved in polynomial time (such as integer factorization, or discrete logarithm).

Let's look at RSA:
you choose your $K_{pub} = (n,e)$ and $k_{priv} = (d)$ with $e \in \{1,2,\dotsc , \Phi(n) - 1\}$ to fulfill  the following equations:

$$\Phi(n) = (p - 1) \cdot (q - 1)$$
$$\operatorname{gcd}(e, \Phi(n)) = 1$$
$$d \cdot e \equiv 1 \bmod \Phi(n)$$

Every element in a group can have one inverse element at maximum, you can not find a $d'$ for which:

$$d' \cdot e \not\equiv 1 \bmod \Phi(n)$$

So instead of:

$$d_{k_{priv}}(y) = d_{k_{priv}}(e_{k_{pub}}(x)) = (x^e)^d \equiv x^{de} \equiv x \bmod n$$

you will compute:

$$d_{k'_{priv}}(y) = d_{k'_{priv}}(e_{k_{pub}(x)}) = (x^e)^{d'} = x^{d'e} \equiv m' \not\equiv m \bmod n$$

So you will be able to compute the decryption with a different $d'$, but your result $m'$ will have nothing in common with the original $m$.