I am not an expert regarding quantum channels etc., but I think this is a pure math question. Let's assume that we're working on a finite-dimensional complex Hilbert space.
In finite dimensions, the defining properties of a density matrix $\rho$ are positive semi-definiteness $\rho \geq 0$ and the trace-normalization $\mathrm{Tr}\,\rho=1$. Properties $1$ and $4$ are consequences of this definition. In particular, a positive semi-definite operator is hermitian$^\dagger$.
To answer your first question: Let $\phi$ denote a map which maps positive semi-definite operators to positive semi-definite operators. Following the considerations above, we see that if $\rho$ is positive semi-definite, then $\phi(\rho)$ is hermitian.
I don't know if I correctly understand your second question, but note that Kraus operators preserve the normalization and positive semi-definiteness. To see this, define for a density operator $\rho$ the following map: $$\varphi(\rho):= \sum\limits_k B_k \,\rho \, B_k^\dagger \quad , $$
with $\displaystyle \sum\limits_k B_k^\dagger\, B_k = \mathbb I$, where $\mathbb I$ denotes the identity operator. We find
$$ \mathrm{Tr}\,\varphi(\rho) = \mathrm{Tr}\, \sum\limits_k B_k\, \rho\,B_k^\dagger = \mathrm{Tr}\, \sum\limits_k B_k^\dagger B_k\, \rho = \mathrm{Tr}\, \mathbb I\,\rho = \mathrm{Tr}\,\rho = 1 \quad , $$
where we've used the linearity and the cyclic property of the trace. Moreover:
$$\langle \psi|\varphi(\rho)|\psi\rangle = \sum\limits_k \langle \psi|B_k\, \rho \,B_k^\dagger|\psi\rangle = \sum\limits_k \langle \underbrace{\psi_k|\rho|\psi_k\rangle}_{\geq 0} \geq 0 \quad , $$
with $|\psi_k\rangle := B_k^\dagger|\psi\rangle$.
So $\varphi(\rho)$ is a density matrix again and therefore pure if and only if it is a projector. However, note that Kraus operators can change the purity of a density operator, cf. this quantum computing question.
$^\dagger$ This is easy to see: For a positive semi-definite $\rho$, define the following hermitian operators:
\begin{align}
\rho_1 &:= \frac{1}{2} (\rho+ \rho^\dagger) \\
\rho_2 &:= \frac{1}{2i} (\rho -\rho^\dagger) \quad .
\end{align}
Then obviously $\rho = \rho_1 + i\,\rho_2$ and since $\langle \psi |\rho|\psi\rangle , \langle \psi |\rho_1|\psi\rangle , \langle \psi |\rho_2|\psi\rangle \in \mathbb R$, it follows that $\langle \psi|\rho_2|\psi\rangle \overset{!}{=}0$ for all $|\psi\rangle$. Thus $\rho_2=0$ and $\rho=\rho_1$, i.e. $\rho$ is hermitian.
