It is well-known that if $G$ is a finite $p$-group acting on a non-zero $\mathbb{F}_p$-vector space $V$, then $V^G \neq \{0\}$.
My question is about a generalization of this result when $G = V = \mathbb{F}_p^\mathbb{N}$ (no topology involved).
(A counter-example with $G = \mathbb{F}_p^{(\mathbb{N})}$ would provide a negative answer since the latter is a direct factor of $\mathbb{F}_p^\mathbb{N}$.)
For every group $G$, the left action $g\cdot f(x)=f(g^{-1}x)$ yields an action on the space $\mathbf{F}_p^{(G)}$ of finitely supported functions $G\to \mathbf{F}_p$. If $G$ is infinite, this action has no nonzero fixed point.
If $G=\mathbf{F}_p^{(\alpha)}$ with $\alpha$ infinite, $G$ itself has cardinal $\alpha$ and hence $\mathbf{F}_p^{(G)}$ is isomorphic to $G$. Hence $G$ has an action on $G$ by group automorphisms, with no nonzero fixed point. This applies to your particular case $\alpha=2^{\aleph_0}$.
Added: in a less specific context: the argument shows that:
For every infinite group $G$, for every field $K$, and cardinal $\alpha\ge |G|$, there exists a representation of $G$ in $K^{(\alpha)}$ with no nonzero finite-dimensional subrepresentation.
The above shows it for $\alpha=|G|$, and the general case follows by taking direct sums of copies of such a representation.
