Note: the following answer is incomplete. Maybe it helps though.
Suppose $\sigma(t)=t$ with $t \in \mathbb{F}_3[[T]]$ an element of $T$-adic valuation $k$ such that $3 \nmid k$. Then I claim that the order of $\sigma$ divides $k$. Indeed, there exists an automorphism $\alpha$ of $R = \mathbb{F}_3[[T]]$ 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.