For a singular n-simplex $\alpha$ and a permutation $t\in S_{n}$, define $t\alpha$ to be the simplex with vertices permuted by $t$. Do we have $t\alpha$ homologous to $\text{sgn}(t)\alpha$ ? And does this gives an action of $S_n$ on $H_n$?
I can see for $n=1$ this is true, since the reverse of vertices gives opposite path in fundamental group, and $H_1$ is the abelianization of it.
A problem with this construction is that the vertex permutation operation does not necessarily map cycles to cycles. Therefore, it cannot possibly define a map on homology groups.
For example, let's choose $X = \mathbb R^3$ as our topological space, and let $p, q, r, s$ be any four distinct points in $X$. Let $\tau \in C_2(X)$ be the chain $$\tau = [q, r, s] - [p, r, s] + [p, q, s] - [p, q, r],$$ where, for example, the singular $2$-simplex $[q, r, s]$ is defined to be the affine map from the standard $2$-simplex to the triangle with vertices $q, r, s$.
Applying the boundary map $\partial$ to $\tau$, we obtain \begin{align} \partial (\tau) & = [r, s] - [q, s] + [q, r] - [r, s] + [p, s] - [p, r] + [q, s] - [p, s] + [p, q] - [q, r] + [p, r] - [p, q] \\ &= 0. \end{align} This shows that $\tau$ is a cycle.
Now, take $t \in S_3$ to be the permutation $t = (1, 2, 3)$. Acting on $\tau$ with $t$ by vertex permutation, we obtain $$ t \tau = [r, s, q] - [r, s, p] + [q, s, p] - [q, r, p].$$
And now, if we apply the boundary map $\partial$ to $t \tau$, we get \begin{align} \partial (t \tau) & = [s, q] - [r, q] + [r, s] - [s, p] + [r, p] - [r, s] + [s, p] - [q, p] + [q, s] - [r, p] + [q, p] - [q, r] \\ & = [s, q] + [q, s] - [r, q] - [q, r], \end{align} which does not vanish. Thus $t \tau$ is not a cycle.
For completeness, I'll answer your original question too.
If $\alpha$ is a singular $n$-simplex, then is $t\alpha$ homologous to $\alpha$?
In other words:
If $\alpha$ is a singular $n$-simplex, then does there exist a singular $(n+1)$-chain $\rho$ such that $t\alpha - \alpha = \partial \rho$?
Suppose that the statement is true for all singular $n$-simplexes $\alpha$. Then the statement must be true for all singular $n$-chains as well.
Let's consider the example of the singular $2$-chain $\tau$ on $X = \mathbb R^3$, as defined above. Notice that $$ \tau = \partial ([p, q, r, s]).$$
Suppose there exists a singular $3$-chain $\rho$ such that $$t\tau - \tau = \partial \rho.$$
Then it would follow that $$ t\tau = \partial([p, q, r, s] + \rho).$$
Therefore, $$ \partial(t\tau) =\partial \partial([p, q, r, s] + \rho) = 0,$$ since the boundary of a boundary is always zero.
This contradicts our earlier calculation, where we found that $\partial(t\tau)$ is non-zero.
