= Solution
Put $X=1+T$ and $Y=X^p-1$, so that $\phi(f)=f(Y)$. The essential integrality statement for the <Frobenius substitution on cyclotomic power series> is the decomposition into a <finite free module>:
$$
R=\bigoplus_{i=0}^{p-1}X^i\phi(R).
$$
To prove it, reduce modulo $p$. Since $Y\equiv T^p$, grouping the coefficients of a <formal power series> according to their exponents modulo $p$ gives the unique expression $\sum_{j=0}^{p-1}T^jh_j(T^p)$. The change from $1,T,\ldots,T^{p-1}$ to $1,X,\ldots,X^{p-1}$ is a triangular <matrix> with diagonal entries $1$, so this is also a <basis> over $\mathbb F_p[[T^p]]$. Lift these $p$ coefficient series to $R$, subtract their contribution from $f$, divide the remaining error by $p$, and repeat. The sums of the successive lifts converge in the finer of the <topologies on integral formal power series>, because $R$ is complete in that topology. This proves existence of the decomposition over $R$. A relation among its summands reduces to the zero relation modulo $p$, forcing each coefficient series to be divisible by $p$. Repetition forces divisibility by every power of $p$, hence every coefficient series is zero. This proves uniqueness and, in particular, <injectivity> of $\phi$.
For $f=\sum_{i=0}^{p-1}X^i\phi(f_i)$, define
$$
\boxed{S(f)=f_0.}
$$
This is a $\mathbb Z_p$-<linear map>. To calculate the average over <roots of unity>, work over $\mathcal O=\mathbb Z_p[\zeta_p]$. The substitutions $T\mapsto\zeta X-1$ are legitimate: their constant terms are topologically nilpotent, so the defining coefficient sums converge in $\mathcal O$. Since $(\zeta X)^p=X^p$, each $\phi(f_i)$ is unchanged by the substitution. The sums of the powers of the <roots of unity> are $p$ for $i=0$ and zero for $1\leq i<p$. Consequently
$$
\sum_{\zeta\in\mu_p}f(\zeta X-1)=p\phi(f_0).
$$
Thus the average is integral and belongs to $\phi(R)$, and the constructed <Coleman trace operator> has the required property. <Injectivity> of $\phi$ forces uniqueness of any map satisfying that property. Finally, the decomposition of $\phi(g)$ has $f_0=g$ and all other coefficients zero, giving \b[$S\circ\phi=\mathrm{id}_R$].
We will also use the <trace> of multiplication by $h$ on this <finite free module>. After extending scalars to split the conjugates, that <trace> is the sum just computed. Hence
$$
\operatorname{Tr}_{R/\phi(R)}(h)=p\phi(S(h)).
$$
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