Assume that $C$ is a coalgebra with comultiplication $\Delta:C \to C\otimes C$. The higher order comultiplication can be defined inductively as follows(with some abuse of notations we denote them by $\Delta$, again):
$$\Delta:C\to C\otimes C \otimes C\\ \text{with}\\ \Delta:=(\Delta \otimes id) \circ \Delta$$....etc.
Consider the polynomial coalgebra $C=\mathbb{C}[x]$ with divided power structure Then every power of the differentiation operator satisfies $$(\overbrace{T\otimes T \ldots\otimes T}^{n-times} )\circ \Delta=\Delta \circ T^{n}\\ \forall n\in \mathbb{N}$$. See this post
Is there another example of an operator on polynomial coalgebra which satisfies the above equation for all natural number $n$? Is there an example of an operator on polynomial coalgebra which satisfies the above equation for some $n>2$ but does not satisfy for $n=2$?
The dual of the coalgebra $C$ is a power series algebra $C^\vee=\mathbb{C}[[t]]$, where $x^n$ is dual to $t^n/n!$. Your condition on a map $T$ is then equivalent to the following condition on the dual map $S$: $$\prod_{i=1}^n S(f_i)=S^n\left(\prod_{i=1}^n f_i\right)$$ for any $f_1,\dots,f_n\in C^\vee$. In particular, for any $g\in C^\vee$, the map $S(f)=fg$ clearly satisfies this. If $g=\sum a_n t^n$, then this $S$ is the dual of the map $T:C\to C$ given by $T=\sum a_n D^n$, where $D$ is differentiation (note that this sum might be infinite, but it yields a finite result when applied to any particular element of $C$).
Another such $S$ is the map that takes $f(t)$ to $af(0)$ for some fixed $a\in\mathbb{C}$. This is dual to the map $T:C\to C$ which sends $f(x)$ to $af(0)$.
I claim that these are no other examples of such maps $T$. To prove this, let $T$ be any such map and $S:C^\vee\to C^\vee$ be its dual. Then $S$ satisfies the identities above and is continuous with respect to the $t$-adic topology. Note that for any $f$, we have $S(1)S(f)=S(S(f))$, and it follows that $S(f)^n=S^n(f^n)=S(1)^{n-1}S(f^n)$. If $S(1)=0$, this implies $S=0$, so we may assume $S(1)\neq 0$. We have that $S(1)^{n-1}$ divides $S(f)^n$ for all $n$, and it follows that $S(1)$ divides $S(f)$. In particular, setting $f=t$, we can write $S(t)=S(1)h$ for some $h\in C^\vee$. We then have $S(t^n)=S(1)h^n$ for each $n$. It now suffices to show that either $h=t$ (in which case we have $S(f)=S(1)f$ for all $f$ by continuity) or $h=0$ (in which case it is easy to see $S(1)$ must be a constant and $S(f)=S(1)f(0)$ for all $f$ by continuity).
So suppose $h\neq 0$; we wish to show that $h=t$. Let us write $g=S(1)$. By linearity and continuity, $S(f)=g\cdot (f\circ h)$ for all $f\in C^\vee$, where $\circ$ denotes composition of formal power series (in particular, $h(0)=0$ in order for this composition to be defined). We then have $$g^2=S(1)S(1)=S(S(1))=S(g)=g\cdot (g\circ h).$$ Cancelling $g$, we find $g=g\circ h$. Now we have $$g^2h=S(1)S(t)=S(S(t))=S(gh)=g\cdot ((gh)\circ h))=g\cdot (g\circ h)\cdot (h\circ h)=g^2\cdot (h\circ h).$$ Cancelling $g^2$ we conclude that $h=h\circ h$. If $n$ is the least power of $n$ that has a nonzero coefficient in $h$, this equation gives $n=n^2$ so $n=1$ (since $h(0)=0$ so $n>0$). But then composition with $h$ is an invertible map $C^\vee\to C^\vee$ (we can solve term-by-term for a power series inverse of $h$ with respect to composition), so $h(t)=h(h(t))$ implies $t=h(t)$, as desired.