This is always possible. Namely, fix a metric $<,>$ on $M$ and an orthonormal basis $X_1,...,X_n$ of $\mathfrak{g}$. Set now $\rho(x)=\sqrt{\sum_i ||d\psi_x(X_i)||^2+1}$ and consider the conformally equivalent metric $<,>'=\frac{1}{\rho(x)}<,>$. Then if $v=\sum_i a_i X_i$ we get $$ ||d\psi_x(v)||^2\le ||(a_1,...,a_n)||^2_{\ell_2}(\sum_i||d\psi_x(X_i)||_{<,>'}^2)= $$ $$ ||v||^2\frac{\sqrt{\rho(x)^2-1}}{\rho(x)}\le||v||^2 $$

This is always possible. Namely, fix a metric $\langle\cdot,\cdot\rangle$ on $M$ and an orthonormal basis $X_1,...,X_n$ of $\mathfrak{g}$. Set now $\rho(x)=\sqrt{\sum_i \|d\psi_x(X_i)\|^2+1}$ and consider the conformally equivalent metric $\langle\cdot,\cdot\rangle'=\frac{1}{\rho(x)}\langle\cdot,\cdot\rangle$. Then if $v=\sum_i a_i X_i$ we get $$ \|d\psi_x(v)\|^2\le \|(a_1,...,a_n)\|^2_{\ell_2}\left(\sum_i\|d\psi_x(X_i)\|_{\langle\cdot,\cdot\rangle'}^2\right) = $$ $$ \|v\|^2\frac{\sqrt{\rho(x)^2-1}}{\rho(x)}\le\|v\|^2. $$