This is algebraic dynamics, a big field that I’m outsider to, though I have made some peripheral contributions to it.

First, I’m assuming that you’re asking about ring automorphisms of $\Bbb Z[[x]]$ and of $\Bbb F_p[[x]]$, and the involutions from among these. Under this assumption, $x\mapsto1+x$ *is not* an involution of $\Bbb F_2[[x]]$, since it wants to take the nonunit $x$ to the unit $1+x$.

Second, and more generally, in view of the fact that all automorphisms of $\Bbb F_q[[x]]$ are necessarily continuous, we don’t need to specify that our involution $x\mapsto u(x)$ have no constant term in $u$.

Third, though Will has described an infinite set of nonconjugate involutions (in the group of all invertible series over $\Bbb F_2$), as I recall, according to a theorem of Klopsch, these are all that there are out there, in this $\Bbb F_2$-case.

Finally, all the involutions of $\Bbb Z[[x]]$ that I know about are (naturally) the $[-1]$-automorphisms of $\Bbb Z$-formal group laws, and these look like $-x+cx^q\pmod{x^{q+1}}$ for a prime power $q$. Surely there are other involutions in this situation, but I don’t know them. In particular, if you try $n=5$ here, you get a series $x+x^6+\cdots\in\Bbb F_2[[x]]$ that doesn’t seem to come from $[-1]_F(x)\in\Bbb Z[[x]]$ for any $\Bbb Z$-formal group law $F$. But globally-defined formal group laws I’m rather ignorant about. Nobody’s saying, either, that a $\Bbb Z$-involution has to belong to a formal group over $\Bbb Z$.

I think your question has to remain unanswered, at least with the expertise exhibited here so far.

**EDIT — Addendum**:

All thanks to Will Sawin for his persistence in the face of my repeated careless computational errors. The principle involved is that for an integral domain $R$, the map on the semigroup of formal series under substitution $u(x)\in R[[x]]$ with no constant term and $u'(0)\ne0$, namely
$$
u(x)=xw(x)\mapsto \bigl(u(x^m))^{1/m}=x\bigl(w(x^m)\bigl)^{1/m}\,,
$$
is a homomorphism, as long as it’s defined. For definedness, you need $u'(0)=w(0)$ to have an $m$-th root in the base, and more broadly, you need some sort of convergence theorem to tell you that you can take the $m$-th root of the series $w(x)$.

In the comments, I pointed out that $(1+p^2x)^{1/p}\in\Bbb Z[[x]]$, and this means that $(1+m^2x)^{1/m}\in\Bbb Z[[x]]$ as well, for any positive integer $m$.

Will says, start with the fundamental involution $u(x)=-x/(1+x)=\sum_1^\infty(-x)^n$, first apply $u(x)\mapsto u(m^2x)/m^2=U(x)$, and apply the “stretching-out” operation above, which will be defined as long as $U'(0)=-1$ has an $m$-th root in $\Bbb Z$, namely as long as $m$ is odd. Since $U(X^m)=-x^m+m^2x^{2m}-m^4x^{3m}+\cdots=-x^m[1-m^2x^m+m^4x^{2m}-\cdots]$, the conditions for the existence of the $m$-th root are satisfied. Thus $\bigl(U(x^m)\bigr)^{1/m}$ is a good lifting to $\Bbb Z[[x]]$ of the characteristic-two involution of the form $-x(1+g(x^m))$ for $m$ odd.

of$\mathbb F_p[[x]]$. Rather, we are considering involutions $f\in\mathbb F_p[[x]]$. $\endgroup$ – Wojowu May 5 '19 at 20:22in$\mathbb{F}_p[[x]]$, noton$\mathbb{F}_p[[x]]$. Perhaps, that was a source of confusion. $\endgroup$ – Alexander Burstein May 5 '19 at 20:24