= Solution
An <invariant subspace> $W$ satisfies $D(g)W\subseteq W$ for every $g\in G$; applying the inverse gives equality. The representation is <irreducible> when the only invariant subspaces are $0$ and the full space.
The inverse and <conjugate transpose> each reverse multiplication order, so their combination preserves it:
$$
D_2(gh)=[D_1(h^{-1})D_1(g^{-1})]^\dagger=D_1(g^{-1})^\dagger D_1(h^{-1})^\dagger=D_2(g)D_2(h).
$$
Its values are invertible, so this is the <conjugate contragredient representation>.
If $W$ is invariant under $D_2$, take $w\in W$ and $v\in W^\perp$. For the standard <Hermitian inner product>,
$$
\langle w,D_1(g)v\rangle=\langle D_1(g)^\dagger w,v\rangle=\langle D_2(g^{-1})w,v\rangle=0,
$$
since $D_2(g^{-1})w\in W$. Hence $D_1(g)v\in W^\perp$, proving its invariance. If $D_1$ is irreducible, $W^\perp$ is either zero or the whole space; finite-dimensional orthogonal complementation makes $W$ the whole space or zero, respectively. Thus \b[$D_2$ is irreducible whenever $D_1$ is irreducible].
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