How to get the relation for dependence of anomalous dimension on regularization? Here is the anomalous dimension:
$$
\gamma_{\Gamma}(t, g) = \left[\frac{\partial }{\partial t}\ln \left(Z_{\Gamma}(t , g) \right)\right]_{t = 1},
$$
where $Z_{\Gamma}$ is renormalization factor which arises in n-point functions $\Gamma $, $t$ denotes change of renormalization parameter $t = \frac{\mu{'}}{\mu}$. $Z_{\Gamma}$ arises explicitly after making shift of renormalization parameter $\mu$ (for fixed type of renormalization):
$$
\Gamma (xt , g) = Z_{\Gamma}^{-1}(t , g) \Gamma (x, \bar{g}(t , g)), \quad x = \frac{k}{\mu}, \quad t = \frac{\mu}{\mu{'}}.
$$
Let's change type of regularization (coupling constant will change to $g \to \tilde {g}(g)$. Then n-point function will change as
$$
\Gamma \left(\frac{k}{\mu} , g \right) = q(g) \tilde {\Gamma}\left( \frac{k}{\mu} , \tilde {g}(g) \right).
$$ 
How to get from these equations that $\gamma_{\Gamma}$ will change to
$$
\tilde{\gamma}_{\Gamma}(\tilde {g}(g)) = \gamma_{\Gamma}(g) - \beta (g)\frac{d\ln (q(g))}{dg}
$$
(the definition for $\beta$-function see here)?
 A: I'm not quite sure where the details of the last equations come from, but I think that the step that you are missing is to identify,
$$q(g) = \frac{1}{Z_\Gamma(g)} \, .$$
A: Two functions, $\Gamma \left(\frac{k}{\mu}, g\right)$ and $\Gamma \left( \frac{k}{\mu}, \tilde{g}(g)\right)$, satisfy so called Callan–Symanzik equations:
$$
\tag 1 \left(t\partial_{t} - \beta (g) \partial_{\beta} + \gamma_{\Gamma}(g)\right)\Gamma (t , g) = 0, \quad \left(t\partial_{t} - \tilde{\beta} (\tilde{g}) \partial_{\tilde{\beta}} + \tilde{\gamma}_{\Gamma}(\tilde{g})\right)\tilde{\Gamma} (t , \tilde{g}) = 0.
$$
Let's use identities (the second one is the result of the definition of $\beta$-function) 
$$
\Gamma(t, g) = q(g)\tilde{\Gamma}(t, \tilde{g}), \quad \tilde{\beta}(\tilde{g}) = \frac{d \tilde {g}}{dg}\beta (g)
$$
and let's insert them into the second equation of $(1)$ with using the first one:
$$
\left(t\partial_{t} - \beta (g) \partial_{\beta} + \tilde{\gamma}_{\Gamma}(\tilde{g})\right)\frac{1}{q(g)}\Gamma (t , g) = \frac{1}{q(g)}\left( t\partial_{t} - \beta (g) \partial_{\beta} + \gamma_{\Gamma}(g) \right) +
$$
$$
+ \Gamma (t ,g)\left[ \frac{1}{q(g)}\tilde{\gamma}_{\Gamma}(\tilde{g}) - \frac{1}{q(g)}\gamma_{\Gamma}(g)-\beta (g)\partial_{g}\left( \frac{1}{q(g)} \right)  \right] = 
$$
$$
=\Gamma (t ,g)\left[ \frac{1}{q(g)}\tilde{\gamma}_{\Gamma}(\tilde{g}) - \frac{1}{q(g)}\gamma_{\Gamma}(g)-\beta (g)\partial_{g}\left( \frac{1}{q(g)} \right)  \right]= 0 \Rightarrow
$$
$$
\tilde{\gamma}_{\Gamma}(\tilde{g}) = \gamma_{\Gamma}(g) - q(g) \beta (g)\frac{q'(g)}{q^{2}(g)} = \gamma_{\Gamma}(g) - \beta(g)\partial_{g}ln (q(g)).
$$
