If $G$ is finite over $k$, then $\mu$ is automatically finite and flat. Indeed the morphism $G\to\mathrm{Spec}\;k$ is finite and flat and hence so is the projection $p:G\times_k X\to X$. But $\mu$ only differs from $p$ by an automorphism of $G\times_k X$ and hence it is also finite and flat. Namely $\mu=p\circ \alpha$ for $\alpha(g,x)=(g,gx)$.
Let me add the remark that more generally for a groupoid $X_1 \overset{d_1}{\underset{d_0}{\rightrightarrows}} X_0$ as in SGA3.V there is an automorphism $\alpha$ of $X_1$ such that $d_1=d_0\circ \alpha$. This removes the a priori strange asymmetry in for example Theorem 4.1 of loc.cit., where only the map $d_1$ is required to be finite locally free.
If $G$ is finite over $k$, then $\mu$ is automatically finite and flat. Indeed the morphism $G\to\mathrm{Spec} k$ G\to\mathrm{Spec}\;k$is finite and flat and hence so is the projection$p:G\times_k X\to X$. But$\mu$only differs from$p$by an automorphism of$G\times_k X$and hence it is also finite and flat. Namely$\mu=p\circ \alpha$for$\alpha(g,x)=(g,gx)$. 1 If$G$is finite over$k$, then$\mu$is automatically finite and flat. Indeed the morphism$G\to\mathrm{Spec} k$is finite and flat and hence so is the projection$p:G\times_k X\to X$. But$\mu$only differs from$p$by an automorphism of$G\times_k X$and hence it is also finite and flat. Namely$\mu=p\circ \alpha$for$\alpha(g,x)=(g,gx)\$.