Encrypted hash vs hashed encryption
What you want does not exists with any secure encryption scheme and collision resistant hash function.

Any encryption scheme that is at least CPA secure¹ cannot be deterministic. Therefore not only can the equality you ask for not hold, it's even true that for $c_1 = \mathsf{Enc}(k,H(m))$ and $c_2 = \mathsf{Enc}(k,H(m))$, it holds with overwhelming probability that
$c_1\neq c_2$. similarly, $h_1\neq h_2$ with overwhelming probability for $h_1 = H(\mathsf{Enc}(k,m))$ and $h_2 = H(\mathsf{Enc}(k,m))$, since the encryption scheme is secure and the hash function is collision resistant.

Furthermore, due to a secure encryption scheme being probabilistic, ist must also be expanding. I.e., for $c = \mathsf{Enc}(k,m)$ it holds that $|c|&gt;|m|$.

Since a hash function compresses inputs to a fixed length, it must therefore hold that $$|\mathsf{Enc}(k,H(m))| &gt; |H(\mathsf{Enc}(k,m))|$$
and thus by extension necessarily that
$$\mathsf{Enc}(k,H(m))\neq H(\mathsf{Enc}(k,m))$$
even if you would try to somehow correlate the randomness used in the two encryptions.



¹This is the absolute bare minimum of what we consider a secure encryption scheme in most cases. As Maarten points out in the comments there are some narrowly defined use-cases in which weaker notions of security may be acceptable.