Is the Rice's theorem applicable to $\{ \langle M \rangle \mid M \mbox{ is a Turing machine such that }L(M) = H_{all} \mbox{ } \}$?
Well, Rice's theorem doesn't apply but we don't need it&mdash;$L^*$ is empty and therefore decidable. 

To figure this out, we just need to be meticulous about what these languages are.


$L^*$ is the set of all Turing machines that recognize the totality problem. It is the set of all Turing machines that can take in a Turing machine $\langle N\rangle$ and accept just if $N$ halts on all inputs.
$L^\prime$ is the set of all Turing machines that halt on all inputs.
They're different languages: $L^*$ is the set of all machines whose language is $L^\prime$.


The totality (all-halt) problem is unrecognizable: there is no Turing machine that takes in a TM $\langle N\rangle$ and accepts just if $N$ halts on all inputs. (Consider a reduction to the halting problem.) Therefore, $L^*$ is empty and $L^\prime$ is unrecognizable.

$L^*$ is empty, therefore decideable (when given a machine $\langle N\rangle$ and asked if it recognizes the totality problem, always say no.) $L^\prime$ is unrecognizable, which is stronger than merely undecidable.

But, if we didn't already know that $L^\prime$ was unrecognizable, we could use Rice's theorem to show that $L^\prime$ is undecidable. There's a string in $L^\prime$, for example the Turing machine that ignores its input and always accepts. There's a string not in $L^\prime$, for example the Turing machine that ignores its input and loops forever.



Also, a side note: Rice's theorem says that non-trivial language properties of TMs are undecideable. As you point out, there are two trivial languages which the theorem doesn't cover: the empty language $\varnothing$ and the language of all strings $\Sigma^*$. Note that we don't need Rice's theorem for these languages to determine decidability&mdash; we already know they're decideable. In fact, they're regular: there's a DFA that accepts all strings, and a DFA that accepts no strings.