Let $V, W$ be $\mathbb{C}$-vector spaces. Suppose that a finit group $G$ acts on them. Let $\phi :V \rightarrow W$ be a linear map, which induces $G$-invariant linear map $\phi^{G} : V^{G} \rightarrow W^{G}$. Here, $V^G$ is the set of all $G$-invariant vectors in $V$ and $\phi^{G}$ is the restriction of $\phi$ to $V^G$.
Then, my question is
Does $({\mathop{\mathrm{coker}}\nolimits}\ \phi^{G}) = 0$ lead $({\mathop{\mathrm{coker}}\nolimits}\ (\phi))^{G} = 0$? and $(\ker{\phi})^{G} \subset \ker(\phi^{G})$ ?
For the kernel part we have $(\ker\phi)^G=\ker\phi\cap V^G$, which is contained in $\ker(\phi^G)$.
For the cokernel part I am assuming that $\text{image}(\phi)$ is $G$-invariant, since only then we have an action of $G$ on $\text{coker}(\phi)$. (This happens if $\phi$ is $G$-equivariant).
Suppose $w+\text{image}(\phi)\in (\text{coker}\phi)^G$. Then $g\cdot w +\text{image}(\phi) = w+ \text{image}(\phi)$ for all $g\in G$. Therefore $$ \left(\frac{1}{|G|}\sum_{g\in G} g\cdot w\right) +\text{image}(\phi) = w + \text{image}(\phi) $$ The vector $(\sum_{g\in G}g\cdot w)/|G|$ is a $G$-invariant vector, and hence, by assumption that $\text{coker}(\phi^G) = 0$, it lies in $\text{image}\phi^G$. Therefore it also lies in $\text{image}(\phi)$, implying that $w+\text{image}(\phi) = 0$.
So we have shown that an arbitrary vector in $(\text{coker}(\phi))^G$ is zero and hence $(\text{coker}(\phi))^G=0$.