$\DeclareMathOperator\SL{SL}
\DeclareMathOperator\GL{GL}
\DeclareMathOperator\PSL{PSL}
\DeclareMathOperator\Aut{Aut}
\DeclareMathOperator\Out{Out}
\DeclareMathOperator\Ad{Ad}
$I can only address Question 1/Question 4. Let me replace $\SL_n$ with $\GL_n$; our $H$ will be simple, so it will provide an example for $\SL_n$ also.
If $\rho, \psi: H \to \GL_n(\mathbb C)$ are faithful representations with the same image, then $\psi^{-1} \circ \rho$ is an automorphism of $H$. Hence asking whether $(\tau\rho)(H)$ is conjugate to $\rho(H)$ as a set for faithful $\rho$ is asking whether there is an automorphism $\varphi \in \Aut(H)$ such that $\tau \rho$ is isomorphic to $\rho \varphi$ as a complex representation of $H$. This reduces the problem to character theory. Note also that the action of $\Aut(H)$ on characters factors through $\Out(H)$.
An example is given by $H = \PSL_2(\mathbb F_p)$ for $p > 13$.
For any nonsquare $a \in \mathbb F_p^\times$,
$$\Out(H) = \left\langle \Ad \begin{pmatrix} a & 0 \\ 0 & 1\end{pmatrix}\right\rangle \cong \mathbb Z/2\mathbb Z.$$
See [1] or [2] for this.
Let
$$T = \left\{ \begin{pmatrix} x & 0 \\ 0 & x^{-1} \end{pmatrix} \mid x \in \mathbb F_p^\times \right\} / \{\pm 1\}$$
be the diagonal torus in $H$.
If $\zeta$ is a $(p-1)/2$th root of unity, let $\alpha: \mathbb F_p^\times \to \mathbb C^\times$ be a multiplicative character sending a primitive element of $\mathbb F_p^\times$ to $\zeta$. Note that $\alpha(-1) = 1$.
There is an irreducible representation of degree $p+1$ given by parabolically inducing the character $\begin{pmatrix} x & 0 \\ 0 & x^{-1} \end{pmatrix} \mapsto \alpha(x)$ of $T$ up to $H$. Its character $\chi$ satisfies
$$ \chi\begin{pmatrix} x & 0 \\ 0 & x^{-1}\end{pmatrix} = \alpha(x) + \alpha(x^{-1}).$$
See e.g. [3] for a description of the characters of $\mathrm{PSL}_2(\mathbb F_p)$. If $\varphi$ is the automorphism of conjugation by $\begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}$, then $\varphi$ fixes $T$ elementwise, so $\chi\varphi$ has the same values on $T$ as $\chi$.
Lemma: If $\zeta$ is a primitive $m$th root of unity, then $\zeta^{-1} + \zeta$ is irrational for $m \notin \{1,2,3,4,6\}$.
This lemma also appears in the theory of root systems.
If $\zeta$ is chosen to be a primitive $(p-1)/2$th root of unity, then $\zeta + \zeta^{-1}$ is irrational for $(p-1)/2 \notin \{1,2,3,4,6\}$, i.e. $p = 11$ or $p > 13$. This provides an example of a character $\chi$ and $\tau \in \Aut(\mathbb C)$ such that $\tau \chi \neq \chi \varphi$ for all $\varphi \in \Out(H)$.
Proof of the lemma: We may assume $m > 2$.
The degree $[\mathbb Q(\zeta):\mathbb Q]$ is $\phi(m)$.
As $\zeta + \zeta^{-1}$ is real, $[\mathbb Q(\zeta): \mathbb Q(\zeta + \zeta^{-1})] > 1$, but $\zeta$ satisfies the quadratic equation
$$\zeta^2+1=(\zeta+\zeta^{-1})\zeta $$
over $\mathbb Q(\zeta + \zeta^{-1})$, so $[\mathbb Q(\zeta):\mathbb Q(\zeta + \zeta^{-1})] = 2$. Hence the degree of $\zeta^{-1} +\zeta$ is $\phi(m)/2$. If $m = p_1^{k_1}\cdots p_\ell^{k_\ell}$, then $\phi(m) = p_1^{k_1 - 1}(p_1 - 1)\cdots(p_\ell^{k_\ell})(p_\ell - 1)$. If $\phi(m) = 2$, then primes $p_i$ appearing must have $p_i - 1$ divides $2$, whence $p_i$ can be only $2$ or $3$. Further, only $m=3$, $m=4$, and $m=6$ are possible.
References
[1] Steinberg, R. (1960). Automorphisms of Finite Linear Groups. Canadian Journal of Mathematics, 12, 606-615. doi:10.4153/CJM-1960-054-6
[2] What is the outer automorphism group of $\operatorname{SL}(2,\mathbb{F}_q)$?
[3] Bonnafé, Representations of $SL_2(\mathbb F_q)$, §5.3, or Fulton and Harris, Representation Theory, §5.2
