Coalgebras(or quantum groups) which admit a linear operator satisfying certain functional equation What is an example of a coalgebra $C$ which admit a linear operator $T$, different from scalar operators $T=\lambda Id$, which satisfy $$(T\otimes T)\circ \Delta=\Delta \circ T^2 $$ but $C$ is not isomorphic to a subcoalgebra of $\mathbb{C}[x]$ or the trigonometric coalgebra generated by $c=cos, s=sin$?
Note that the differentiation operator on these coalgebras satisfies the above functional equation.
Our next question is the following:
 Does the above functional equation implies the automatic continuity of $T$ when $T$ is a symmetric operator on a quantum group?(By symmetricity of $T$ we mean $T(a^*)={(T(a))}^*$
Edit: According to the comment of Konstantinos, I add this link as a motivation for this question:
http://bims.iranjournals.ir/article_872_fd7287eb7f1365d9156e9da3ccb25196.pdf
 A: About your first question:
Since you are asking for an example, take any group hopf algebra $k\mathbb{G}$, pick some subset $S\subset \mathbb{G}$ and denote  $kS$ the linear subspace of $k\mathbb{G}$ generated by the set $S$.  Let $T:k\mathbb{G}\rightarrow kS\subset k\mathbb{G}$ be the projection operator onto that subspace. Then your property is  satisfied, by direct computation of both sides of your equation: 
$$
\Delta\circ T^2\big(\sum_{g\in\mathbb{G}}k_g g\big)=\sum_{g\in S}k_g g\otimes g=(T\otimes T)\circ\Delta\big(\sum_{g\in\mathbb{G}}k_g g\big)
$$
since $T(g)=g$, if $g\in S\subset \mathbb{G}$ and $T(g)=0$, if $g\in G\setminus S$.  
From a more general point of view, since $(f\otimes f)\circ \Delta=\Delta\circ f$ is -by definition- satisfied for any morphism of coalgebras, then your functional equation should be satisfied for any idempotent ( i.e. $T^2=T$) coalgebra endomorphism $T:C\rightarrow C$.  
Edit: Attempting a translation of the comments to the OP, on the dual relation on algebras, by users : @მამუკა ჯიბლაძე and @user44191:
The functional equation stated at the OP implies that:
a). $\Delta\big(T^2(C)\big)\subseteq T(C)\otimes T(C))$ and
b). $T^2$ acts on $C$ as: 
$$T^2(c)=\varepsilon\big(T^2(c)_1\big)T^2(c)_2=T^2(c)_1\varepsilon\big(T^2(c)_2\big)= \\ =\varepsilon\big(T(c_1)\big)T(c_2)=T(c_1)\varepsilon\big(T(c_2)\big)$$
(you can get that by applying $Id\otimes\varepsilon$ and $\varepsilon\otimes Id$ on both sides of the functional equation at the OP $(T\otimes T)\circ \Delta=\Delta \circ T^2$).   
However, i do not know about your second question in general.
