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For your question 2, the usual proof caries over easily and one can see that $X$ is the space of $\mathbb{H}$-algebra morphism from $H(X)$ to $\mathbb{H}$ hence if $H( X) \simeq H(Y)$ then $X \simeq Y$ and the isomorphism between $H(X)$ and $H(Y)$ is induced b the isomorphism between $X$ and $Y$..

For your question 2, the usual proof caries over easily and one can see that $X$ is the space of $\mathbb{H}$-algebra morphism from $H(X)$ to $\mathbb{H}$ hence if $H( X) \simeq H(Y)$ then $X \simeq Y$ and the isomorphism between $H(X)$ and $H(Y)$ is induced by the homeomorphism between $X$ and $Y$..

But one can give a more explicit condition: A quaternionic $C^*$-algebra is of the form $H(X)$ if and only if its self adjoint element (x^*=x) are central.

Indeed, it is then obvious that the set of self-adjoint element is a real $C^*$-algebra with trivial involution hence that it is of the form $C(X,\mathbb{R})$ for some $\mathbb{R}$.

But one can give a more explicit condition: A quaternionic $C^*$-algebra is of the form $H(X)$ if and only if its self adjoint element ($x^*=x$) are central (or just commute between themselves and to elements of $\mathbb{H}$).

Indeed, it is then obvious that the set of self-adjoint element is a real $C^*$-algebra with trivial involution hence that it is of the form $C(X,\mathbb{R})$ for some $X$.

So I thought a little about those question and here are the partial answer I have been able to find, I'm still thinking about it so I will edit if I found more answer.

Let $A$ be a quaternionic $C^*$-algebra, and $a \in A$ I guess you want to define the spectrum as $\{ h \in \mathbb{H} | a - h \}$ is non-invertible.

'non-empty' is trickier because the spectrum in a real $C^*$-algebra don't have to be inhabited and we don't have holomorphic calculus at disposition for complex $C^*$-algebra. But it seems difficult to produce a counter example: For example, as soon as you have an element which commute to some purely imagnary unit $u$, then $x,x^*,u$ generates a complexe $C^*$-algebra (with $u$ as $i$) hence $x$ has a non-empty spectrum inside of it (and we can hope that it implies that $x$ has a non-empty sepctrum in the initial algebra, although this is not obvious either).

One can do better: A quaternionic $C^*$-algebra is of the form $H(X)$ if and only if "self adjoint element are central".

The only part I don't know how to answer is the non-emptyness of the spectrum... I'm still thinking about it, and it might be related to those possible additional condition I'm mentioning just above. I will edit if I find something.

So, let $A$ be a quaternionic $C^*$-algebra, and $a \in A$ I guess you want to define the spectrum as $\{ h \in \mathbb{H} | a - h \}$ is non-invertible.

As I said, 'non-empty' is trickier because the spectrum in a real $C^*$-algebra don't have to be inhabited and we don't have holomorphic calculus at disposition for quaternionic $C^*$-algebra (althoug it might be an idea to develop it a little, the resolvant is still locally a formal series so maybe one can use a kind of quaternionic Liouville's theorem). But it seems difficult to produce a counter example: For example, as soon as you have an element which commute to some purely imagnary unit $u$, then $x,x^*,u$ generates a complexe $C^*$-algebra (with $u$ as $i$) hence $x$ has a non-empty spectrum inside of it, and it implies that $x$ has a non-empty sepctrum (as in real $C^*$-algebra it also true that $x$ is invertible is a subalgebra if and only if it is invertible in the larger algebra: it is proved by taking the complex form of the algebras).

But one can give a more explicit condition: A quaternionic $C^*$-algebra is of the form $H(X)$ if and only if its self adjoint element (x^*=x) are central.