Convolution algebra as a bialgebra? 
*

*Context
Let $(A, \mu, \eta, \Delta, \epsilon)$ be a bialgebra over a field $k$. Consider the vector space $\mathrm{End}(A)$ over $k$.
Define the convolution product $$*: \mathrm{End}(A)\otimes \mathrm{End}(A) \rightarrow \mathrm{End}(A); \qquad f \otimes g \mapsto \mu \circ (f \otimes g)\circ \Delta.$$
Define the unit map $$\overline \eta: k \rightarrow \mathrm{End}(A); \qquad 1 \mapsto  \eta \circ \epsilon.$$ 
Then $(\mathrm{End}(A), *, \overline \eta)$ becomes an associative, unital algebra.

*Questions


*

*Can $(\mathrm{End}(A), *, \overline \eta)$ be made into a bialgebra?  

*Does it become a Hopf algebra that way?  

*Is there a canonical way?


 A: This is not a complete answer, but it is too long to be a comment.
Let me answer your questions in the finite-dimensional case.
(1) Assume that $A$ is finite-dimensional as a vector space. Then, as vector spaces,
$$
\begin{array}{ccc}
\mathrm{End}(A) & \cong & A \otimes A^* \\
f & \to & \sum_if(e_i) \otimes e_i^* \\
\left[b\mapsto a\varphi(b)\right] & \leftarrow & a \otimes \varphi
\end{array}$$
where $\{e_i\}$ is a basis of $A$ and $\{e_i^*\}$ is the corresponding dual basis of $A^*.$
(2) The construction of the algebra structure can be replicated for every $\mathrm{Hom}(C,A)$ where $A$ is an algebra and $C$ a coalgebra. In particular, $A^*$ always admits an algebra structure: $(\varphi*\psi)(a) = \sum \varphi(a_1)\psi(a_2)$.
(3) Being $A$ finite-dimensional, $A^*$ admits in fact a bialgebra structure, where $\Delta_*(\varphi) = \sum \varphi_1 \otimes \varphi_2$ is uniquely determined by the rule
$$\sum \varphi_1(a)\varphi_2(b) = \varphi(ab)$$
for all $a,b \in A$ and $\varepsilon_*(\varphi) = \varphi(1)$.
(4) Since both $A$ and $A^*$ are bialgebras, $A \otimes A^*$ is a bialgebra as well with
$$
(a \otimes \varphi)(b \otimes \psi) = ab \otimes \varphi * \psi, \\
\Delta_{A \otimes A^*}(a \otimes \varphi) = \sum \left(a_1 \otimes \varphi_1\right) \otimes \left(a_2 \otimes \varphi_2\right), \\
u_{A \otimes A^*} = u_A \otimes u_{A^*},\\
\varepsilon_{A \otimes A^*} = \varepsilon_A \otimes \varepsilon_{A^*}.
$$
(5) If you consider the algebra structure on $\mathrm{End}(A)$ you gave above and the foregoing algebra structure on $A \otimes A^*$, you will realize that $\mathrm{End}(A) \cong A \otimes A^*$ as algebras. In particular, if you transfer the coalgebra structure on $\mathrm{End}(A)$ then you obtain a bialgebra structure. Therefore, in this case the answer to your first and third question is yes.
(6) The answer to the second question instead is: with the above construction, in general no, unless $A$ is already a Hopf algebra. Assume that you manage to endow $\mathrm{End}(A)$ with an antipode $S_E$. Consider the composition
$$S:= \left(A^* \xrightarrow{1\otimes A^*} A \otimes A^* \xrightarrow{S_E} A \otimes A^* \xrightarrow{\varepsilon \otimes A^*} A^*\right).$$
It satisfies
$$S(\varphi_1)*\varphi_2 = (\varepsilon \otimes A^*)\left(S_E(1 \otimes \varphi_1)(1 \otimes \varphi_2)\right) = (\varepsilon\otimes A^*)(1_A \otimes \varepsilon_*(\varphi)1_{A^*}) = \varepsilon_*(\varphi)1_{A^*}$$
and analogously on the other side (and you may perform the same construction for $A$). Thus you have an antipode on $A^*$ and on $A$.
For the infinite dimensional case, I would say that the answer is no (at least, not "canonically"), but I don't have any counter example to exhibit presently.
