Operator mixing in dimensional regularization of EFTs When renormalizing "non-renormalizable" operators within an effective field theory (EFT) one usually has to introduce additional (higher-dimensional) operators to the Lagrangian which act as counter-terms.
Using dimensional regularization M. Neubert claims in his lectures that one can generally write the bare operators $\mathcal{O}_i^{(n)}$ at mass dimension $n$ in terms of renormalized operators $\mathcal{O}_j^{(n)}(\mu)$ as
$$\mathcal{O}_i^{(n)}=\sum_j Z_{ij}(\mu)\mathcal{O}_j^{(n)}(\mu)$$
where $Z_{ij}$ are the corresponding renormalization factors. This generally means that the resulting renormalization group equations (RGEs) will mix different operators within certain mass dimension but they will not mix operators of different mass dimensions.
Sadly he does not explain this statement any further. However, I would like to understand where it comes from (it might be that it's some obvious reason which I don't see...)
My question is therefore:
Why does dimensional regularization not lead to mixing between operators of different dimension?

What I tried so far: (This might be irrelevant...)
To solve this I tried to start with a "simple" case:
Let's take the example from pages 389-390 from M. D. Schwartz' "Quantum Field Theory and the Standard Model" of a non-renormalizable scalar $\Phi^4$ theory in $d=4$ dimensions
$$\mathcal{L}=-\frac{1}{2}\Phi(\square+m^2)\Phi+\frac{g}{4!}\Phi^2\square\Phi^2$$
Here $g$ has mass dimension $-2$ and following Schwartz the transition amplitude $2\Phi\rightarrow 2\Phi$ at one loop using cut-off regularization i.e.
$$\int_0^\infty dp\rightarrow\int_0^\Lambda dp$$
is proportional to something like
$$A\sim gp^2 \qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad\qquad\;\,\text{(tree-level contribution)}
\\+g^2(C_1\Lambda^4+C_2p^2\Lambda^2+C_3p^4\log(\Lambda)+...)\qquad\text{(1-loop contribution)}$$
where the $...$ denote terms that vanish when taking the limit $\Lambda\rightarrow\infty$. Now, to cancell the divergences that result from the loop contributions one has to introduce additional counterterms to the Lagrangian of the form
$$\lambda \Phi^4$$
which cancels the $\Lambda^4$ divergence and
$$\kappa\Phi^2\square^2\Phi^2$$
which cancels the $\log(\Lambda)$ divergence. Obviously, this would mix operators of different dimensions.
While I don't really see which kind of loop diagram would result in a $p^2$ or $p^0$ dependence (as the $p$ dependence here comes from the vertices and at 1-loop level there will be at least two of them each giving $p^2$), following Schwartz we would still have to add the $\Phi^2\square^2\Phi^2$ term and hence mix operators of dimension 6 and 8.
I don't see how dimensional regularization
$$\int d^dp\rightarrow \mu^{2\epsilon}\int d^{d-2\epsilon}p$$
would change anything here.
 A: tldr: Neuberts statement is only true in massless EFTs when neglecting higher order contributions (i.e. Diagrams with two or more EFT vertices).

As you found out, this statement of Neubert is in general incorrect. In fact it is not even really true for massless EFTs. In my opinion the setup of Operator renormalization is really rather confusing and opaque, especially If one already knows about the renormalization procedure for "normal" QFTs (by this i mean renormalizable theories like QED, etc.) as discussed in pretty much all Textbook on the subject.
EFTs are not fundamentally different from such theories! The only difference between renormalizable and non-renormalizable theories is, that in the latter case we cannot absorb all appearing divergences into a finite set of parameters. The entire renormalization procedure is however exactly analogous. Say we start with a general EFT
$$
\mathcal{L} = \sum_{i,n}C_i^{(n)}\mathcal{O}_i^{(n)}
$$
Where $\mathcal{O}_i^{(n)}$ denote dim-n composite fields and $C_i^{(n)}$ the corresponding
(collection of) coupling parameters (e.g $\mathcal{O}_m^{(3)}=\bar\psi \psi$ and $C_m^{(3)} = m$ or $\mathcal{O}^{(6)}_g=\Phi^2\partial^2\Phi^2$ and $C_g^{(6)} = g / 4!$ and so on). In order to renormalize this theory (in dim-reg. as Neubert "required") we split the parameter in a scale dependent way
$$
C_i^{(n)} = C_i^{(n)R}(\mu) + \delta C_i^{(n)R}(\mu)
$$
(Also one usually factors our some appropriate power of $\mu$ to obtain renormalized parameters of integer dimension). Any Feynman diagram of the theory will give some function in the renormalized parameters $C_i^{(n)R}$, however, in particular the (subrenormalized) divergent part will always be a power series in these variables.
This gives the following "simple" form the the renormalization constants (in MS scheme)
$$
\delta C_i^{(n)R}(\mu) = \sum_{r=2}^\infty Z_{ij_1,...j_r}C_{j_1}^{(n_1)R}(\mu)...
C_{j_r}^{(n_r)R}(\mu), \qquad \text{s.t.}\quad \forall r: 4-n = \sum_l 4 - n_l \\
Z_{ij_1...j_r} = \sum_{k=1}^\infty Z_{ij_1...j_r}^{[k]}\frac{1}{\epsilon} 
$$
From this we see where Neuberts formulation fails. Note that $[C^{(n)}] = 4-n$, which means if the theory has a mass parameter $[m]=1$, then the above equation allows terms like
$$
\delta C^{(n)R} \sim m^R C^{(n+1)R} + ...
$$
Which is precisely the mixing between different dimensional Operators you found. Furthermore, even if there is no mass parameter we can have terms like
$$
\delta C^{(6)R} \sim C^{(5)R}C^{(5)R}
$$
and so on...
In the spirit of perturbation theory we can however argue, that these higher order terms can be neglected and if one has no mass parameters in the EFT, then the first type of mixing also doesn't occur. Then one can absorb the power series in the dimensionless variables into new coefficients $\hat Z_{ij}$ and find
$$
\delta C_i^{(n)R}(\mu) \approx \hat Z_{ij}(\mu) C_j^{(n)R}(\mu)
$$
Finally the fields in the composite operators also pick up a renormalization factor, such that
$$
\mathcal{O}_i^{(n)} = Z_{\mathcal{O_i}}(\mu) {O}_i^{(n)R}(\mu)
$$
and then in the transformed Lagrangian define something like the renormalized Operators from Neubert. Again, the in my opinion better way to think about it is however in terms of renormalized parameters, since there the meaning of every quantity is completely clear.
