$\DeclareMathOperator\tr{tr}\DeclareMathOperator\Int{Int}\DeclareMathOperator\Cent{Cent}$The OP (@ClarkLyons) gave a lovely [generalisation](https://mathoverflow.net/a/394013) of an [answer](https://math.stackexchange.com/a/4153721) of @diracdeltafunk over on MSE.  I believe that the technique can be pushed still further to handle maps to *any* group $H$, as in the original question.  The argument has no new ideas from me.

Suppose that $\phi : G \to H$ is any homomorphism.  Write $\mathscr O_G$ for the set of conjugacy classes in $G$, and $f : H \to \mathbb Z_{\ge 0}$ for the function given by $f(h) = \#\{\gamma \in \mathscr O_G \mathrel: \text{$\phi(g)$ is conjugate to $h$ for all $g \in \gamma$}\}$.

Write $\mathscr O_H$ for the set of conjugacy classes in $H$ and $\hat H$ for the set of irreducible, complex representations of $H$. Define $g : \hat H \to \mathbb C$ by $g(\rho) = \sum_{h \in \mathscr O_H} f(h)\tr \rho(h)$ for all $\rho \in \hat H$.  Then $g(\rho) = \sum_{g \in \mathscr O_G} \tr(\rho \circ \phi)(g)$ is non-negative for all $\rho \in \hat H$.

By orthogonality of the character table, we have that $f(h) = (\#\Cent_H(h))^{-1}\sum_{\rho \in \hat H} g(\rho)\overline{\tr \rho(h)}$ for all $h \in H$, so that
$$
f(h) = \lvert f(h)\rvert \le (\#\Cent_H(h))^{-1}\sum_{\rho \in \hat H} \lvert g(\rho)\rvert\cdot\lvert\tr \rho(h)\rvert \le (\#\eta)(\#H)^{-1}\sum_{\rho \in \hat H} g(\rho)\dim(\rho) = (\#\eta)f(1),
$$
for all $h \in \eta \in \mathscr O_H$.