$\mathrm{Hom}_G(V,W)$ as a subspace of $\mathrm{Hom}(V,W)$

Suppose $V$ and $W$ are both representations of a group $G$, where $V$ and $W$ are $k$-vector spaces. Define $\mathrm{Hom}(V,W)$ to be the space of $k$-linear maps $V \to W$. My notes say that:

$\mathrm{Hom}_G(V,W) = \{ \phi \in \mathrm{Hom}(V,W): g \phi = \phi \}$, and we have a linear projection $\mathrm{Hom}(V,W) \to \mathrm{Hom}_G(V,W)$ given by $\displaystyle \phi \mapsto \frac{1}{|G|} \sum_{g \in G} g \phi$

I'm confused by this. $g \phi = \phi$ isn't enough for $\phi$ to be a $G$-linear map, is it? But yet the projection map fixes those maps that satisfy $g \phi = \phi$, which suggests that this isn't a mistake.

Thanks

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$g \phi$ needs to be understood as the natural action of $G$ on $\text{Hom}(V, W)$, which isn't completely obvious; it sends the map $x \mapsto \phi(x)$ to the map $x \mapsto g \phi(g^{-1} x)$. – Qiaochu Yuan Feb 18 '12 at 18:33
@QiaochuYuan Great, thanks (I'm missing the previous page, which I now assume introduces this action). – Matt Feb 18 '12 at 18:41
Another way of writing the natural action of G on Hom(V,W) is $(g \phi)(g x) = g\,\phi(x)$, which maybe looks "more natural". Or, if instead of writing $\phi(x)$ we write $\langle \phi, x \rangle$ then it becomes $\langle g \phi, g x \rangle = g \langle \phi, x \rangle$ (i.e., "g preserves $\langle \,, \rangle$", although we don't actually have a form here.) – Ted Feb 18 '12 at 18:49
@QiaochuYuan Can you also convert this question to an answer? Again posting to the chat? – Julian Kuelshammer Jun 9 '13 at 19:34

As Qiaochu Yuan remarked in the comments the action of $g$ on homomorphisms is as follows $$(g\phi)(x):=g\phi(g^{-1}x)$$ Now if you have $g\phi=\phi$ then you have $(g\phi)(x)=g\phi(g^{-1}x)=\phi(x)$ for all $x$. Multiplying this by $g^{-1}$ from the left you get $\phi(g^{-1}x)=g^{-1}\phi(x)$. This is obviously the same as $\phi$ being in $\operatorname{Hom}_G(V,W)$.