Note: the following answer is incomplete. Maybe it helps though. Let's write $R=\mathbb{F}_3[[T]]$
Suppose $t \in R$ satisfies $\sigma(t)=t$ and its valuation $k$ is coprime to $3$. Then I claim that the order of $\sigma$ divides $k$. Indeed, there exists (edit: after possibly extending the base field $\mathbb{F}_3$ of $R$, as Michael pointed out below) an automorphism $\alpha$ of $R$ that maps $t$ to $T^k$. So then $\alpha(\sigma(\alpha^{-1}(T)))=\zeta T$ for some $k$-th root of unity $\zeta$, hence $\sigma$ has order dividing $k$.
Conversely, if $\sigma$ has finite order $k$, then symmetric expressions in $T$, $\sigma(T)$, $\ldots$, $\sigma^{(k-1)}(T)$ should give plenty of power series on which $\sigma$ acts trivially.