Let's write $R$ for $\mathbb{F}_3[[T]]$ and $K$ for its fraction field. Suppose $t \in R \setminus \mathbb{F}_3$ satisfies $\sigma(t)=t$. I claim that $\sigma$ then has finite order.
Suppose first that the valuation $k$ of $t$ is coprime to $3$. Then I claim that the order of $\sigma$ divides $k$. Indeed, there exists (after possibly taking a base extension of the ground field $\mathbb{F}_3$, as Michael Zieve pointed out in a comment) 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 $k$.
Now the general case. I am a bit uncertain about this, since it almost seems too easy, but here goes. I first want to prove that if $K_0$ is the topological closure of $\mathbb{F}_3(t)$ in $K$, then $K$ is of finite degree over $K_0$. Now $K_0$ is isomorphic to $\mathbb{F}_3((t))$, hence is a locally compact normed field with respect to the absolute value induced by that on $K$. Furthermore, the field $K=\mathbb{F}_3((T))$ is a locally compact normed $K_0$-vector space, so by e.g. Lemma 2 in these notes by Pete Clark) we conclude that $K$ has finite dimension over $K_0$. But then $\sigma$ extended to $K$ is an element of the finite group $\operatorname{Aut}(K/K_0)$. Therefore $\sigma$ has finite order as an automorphism of $K$, and therefore also as an automorphism of $R$.
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. (This has been worked out much better in Michael's answer and the comments below it.)
