Consider the following embedding of the permutation group $S_4$ inside $S_6$:
$\sigma \in S_4 \to \tilde \sigma \in S_6$, where $$ \tilde \sigma\big(a_{12},a_{13},a_{14},a_{23},a_{24},a_{34}\big)=\big(a_{\sigma(1)\sigma(2)},a_{\sigma(1)\sigma(3)},a_{\sigma(1)\sigma(4)},a_{\sigma(2)\sigma(3)},a_{\sigma(2)\sigma(4)},a_{\sigma(3)\sigma(4)}\big), $$ where we agree that the variables $a_{ij}=a_{ji}$, so for instance if $\sigma(1)=2,\sigma(2)=1$, we think of $a_{\sigma(1)\sigma(2)}$ as $a_{12}$.
Question: Does there exist a continuous function $f:\mathbb{R}^6 \to\mathbb{R}$, such that $f$ is $S_4$-invariant but not $S_6$-invariant?
i.e. I want $$ f\big(\tilde \sigma(a_{12},a_{13},a_{14},a_{23},a_{24},a_{34})\big)=f(a_{12},a_{13},a_{14},a_{23},a_{24},a_{34}) $$ for every $\tilde \sigma \in S_4 \subset S_6$ (and every $a_{ij} \in \mathbb{R}^{6}$), but that
$$ f\big(x_{\tau(1)},x_{\tau(2)},x_{\tau(3)},x_{\tau(4)},x_{\tau(5)},x_{\tau(6)}\big) \neq f(x_1,x_2,x_3,x_4,x_5,x_6), $$
for some $\tau \in S_6\setminus S_4$ and $(x_i) \in\mathbb{R}^{6}$.
I guess we can begin with some non-invariant function, and average it over $S_4$; this probably won't be invariant under the full group $S_6$, generically. Does this work for $$f(a_{12},a_{13},a_{14},a_{23},a_{24},a_{34})=a_{12}+a_{13}?$$
Is there an easy way to compute the averaging of this function? I think the action of $S_4$ on pairs $(i,j)$ is transitive or something like that and this should help.
However, my group theory is rusty, and $S_4$ has $24$ elements...is there a shortcut?