= Solution
The induced <internal hom for modules over a Hopf algebra> on $\operatorname{Hom}_k(H,H)$ is
$$
(h\cdot T)(x)=\sum h_{(1)}T(S(h_{(2)})x),
$$
where both occurrences of $H$ have their left regular action. If $T$ is a morphism of left $H$-<modules>, then
$$
(h\cdot T)(x)=\sum h_{(1)}S(h_{(2)})T(x)=\varepsilon(h)T(x).
$$
So every such morphism is an <invariant vector of a module over a Hopf algebra>.
For the converse, a calculation valid for every $k$-linear $T$ is
$$
\sum(h_{(1)}\cdot T)(h_{(2)}x)=\sum h_{(1)}T(S(h_{(2)})h_{(3)}x)=hT(x).
$$
If $T$ is invariant, its left-hand side is instead $\sum\varepsilon(h_{(1)})T(h_{(2)}x)=T(hx)$. Hence $T(hx)=hT(x)$ for all $h,x$, which is precisely the <R-module homomorphism> condition.
In fact an endomorphism of the left regular <module> is determined by $a=T(1)$ and has $T(x)=xa$. Conversely each right multiplication map is left $H$-linear. \b[Thus the invariant space has the explicit description]
$$
\boxed{\operatorname{Hom}_k(H,H)^H=\operatorname{End}_H(H)=\{R_a:x\mapsto xa\mid a\in H\}.}
$$
As a <vector space> it is isomorphic to $H$, and under composition its algebra is the <opposite algebra> $H^{\mathrm{op}}$. No dimension or semisimplicity hypothesis is needed.
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