The elements in $\mathbb{F}_p[[T]]$ of the form $T+T^2f(T)$ form a group under composition, the Nottingham group $\mathcal{N}_p$. (See https://en.wikipedia.org/wiki/Nottingham_group.) So in the case that $c_1=1$, your $\sigma$ is in $\mathcal{N}_p$, and you're asking if there is some $\tau\in\mathcal{N}_p$ such that $\sigma\star\tau=\tau$. (Here I write $\star$ for the group law, which is composition.) Since $\mathcal{N}_p$ is a group, I can multiply (compose) on the right by $\tau^{-1}$ to conclude that $\sigma(T)=T$. Hmmm... Okay, so I guess this means that your $t$ needs to have a constant term, and also I guess you've assumed that your $f(T)$ is a polynomial, otherwise you can't evaluate $f(t)$ when $t$ is a power series. So this doesn't completely solve your problem, but it at least eliminates all $t$'s that don't have a constant term.
EDIT: As pointed out in the comments, what this argument eliminates is $t$'s of the form $T+T^2g(T)$. (And probably $cT+T^2g(T)$ with $c\ne0$). There remains the interesting question of $t$'s that start with a $T^2$ or higher terms. So my answer is merely a possible start, and also a suggestion that the literature on the Nottingham group might be relevant. The thesis of Matthew Gradner-Spencer might also be relevant: https://repository.library.brown.edu/studio/item/bdr:11323/ He looks at various actions of the Nottingham group on power series starting $T^p+h.o.t.$.