RSA keys are multiplicative inverse in mod phi(n), but also in mod n? I understand that the RSA keys $pk$ and $sk$ are choosen such that one is the multiplicative inverse of the other, in $\mod \phi(n)$

But for the encryption and decryption to work, in other words, that

${m^{pk}}^{sk} \mod n = m$

holds, $pk$ and $sk$ must also be multiplicative inverse to each other in $\mod n$, right?
I understand that the RSA keys $pk$ and $sk$ are chosen such that one is the multiplicative inverse of the other, in $\bmod \phi(n)$


Mostly correct.  Actually, they're always multiplicative inverses modulo $\lambda(n) = \phi(n) / \gcd(n)$; selecting them as inverses modulo $\phi(n)$ does work (they will encrypt and decrypt properly), but using the smaller $\lambda(n)$ also works, and yields smaller key values.


  $pk$ and $sk$ must also be multiplicative inverse to each other in $\bmod n$, right?


That is wrong; $pk \cdot sk \bmod n$ need not be any specific value; there's no reason it needs to be 1.  One trivial example is $n = 85, e = 5, d = 13$; in this case, $e$ is not relatively prime to $n$, hence it doesn't have a multiplicative inverse.

After all, when you look at ${m^{pk}}^{sk} \bmod n$, you're not multiplying by $pk$ or $sk$; you're raising $m$ to the power of $pk, sk$; those are different operations.