Are these sets of indices also index sets?
We start with a bit of notation:


$\phi_x$ is the partial function computed by program $x$.
$W_x = \operatorname{dom} \phi_x$, that is, the set of inputs on which program $x$ halts.
An index set is a set of programs $I$ such that if $\phi_x = \phi_y$ then $x \in I \Leftrightarrow y \in I$. (That is, $I$ is a union of equivalence classes of the equivalence relation $x \equiv y \Leftrightarrow \phi_x = \phi_y$.)


Rice's theorem immediately shows that if $I$ is a non-trivial index sets (that is, $x \in I$ and $y \notin I$ for some $x,y$) then $I$ is not computable.

We can now go over the four sets in the question:


$\emptyset$ is trivially an index set.
$K = \{ e : e \in W_e \}$ is not an index set. Indeed, the recursion theorem shows that some $e$ satisfies $W_e = \{e\}$. There are infinitely many programs which halt only on input $e$ and output $\phi_e(e)$. Pick one of them, $e' \neq e$. Then $e \in K$ and $e' \notin K$ even though $\phi_e = \phi_{e'}$. (This proof is taken from S.&nbsp;Barry Cooper, Computability Theory, mention by Pål GD in the comments.)
$E = \{2,4,6,8,\ldots\}$ is computable and so not an index set by Rice's theorem.
$\{ x : \phi_x = \phi_{x^2} \}$ is (probably) not an index set. Indeed, for it to be an index set it must hold that whenever $\phi_x = \phi_{x^2} = \phi_y$ then also $\phi_y = \phi_{y^2}$, which sounds unlikely. (The recursion theorem gives us infinitely many $x$ such that $\phi_x = \phi_{x^2}$, and for each such $x$, there are infinitely many $y$ such that $\phi_x = \phi_y$. It would be a major coincidence if for all $x,y$ we would have $\phi_y = \phi_{y^2}$.)


In part 4 there is an argument but not a proof, and this is not a coincidence. Whether $I = \{ x : \phi_x = \phi_{x^2} \}$ is an index set or not depends on the definition of $\phi$, that is on the universal Turing machine. Every choice corresponds to an admissible numbering, and there are some admissible numberings for which $I$ is an index set (see here), and there are others for which $I$ is not an index set (see here). So the answer for part 4 is really "we don't know" or "it depends". Nevertheless, for any natural admissible numbering, you will find out that $I$ is not an index set; unless you engineer $I$ to be an index set, it won't be an index set.

For convenience, I will repeat the constructions showing that $I$ can be an index set and can be not an index set:


$I$ can be an index set: Fix some admissible numbering $\phi_x$. Let $p_i$ be the $i$th prime. Define a new admissible numbering $\psi_x$ by $\psi_{p_i^k} = \phi_i$ for all $k \geq 1$, and $\psi_x = 0$ whenever $x$ is not a prime power. Thew new admissible numbering satisfies $\psi_x = \psi_{x^2}$ for all $x$ and so $I = \mathbb{N}$ is trivially an index set.
$I$ can be not an index set: Fix some admissible numbering $\phi_x$. Define a new admissible numbering $\psi_x$ by
$$ \psi_x = \begin{cases} 0 &amp; x = 0,1,2,3 \\ 1 &amp; x = 4 \\ \phi_{x-5} &amp; x \geq 5 \end{cases} $$
In this case $\psi_1 = \psi_{1^2} = \psi_2$ but $\psi_2 \neq \psi_{2^2}$, showing that $I$ is not an index set.


I conjecture that under some reasonable interpretation, the fraction of admissible numberings for which $I$ is an index set tends to zero.