Ramanujan's tau function, $691$ congruence, and $\eta(z)^{12}$

Yes, this is true, as a consequence of an identity in a space of modular forms of weight $6$.

The form $\eta(2z)^{12}$ is in this space; and $\sigma_5(n)$ for $n$ odd are the coefficients of the weight-$6$ form $$ \frac1{1008} \Bigl(E_6(z+\frac12) - E_6(z)\Bigr) = q + 244 q^3 + 3126 q^5 + 16808 q^7 + 59293 q^9 + \cdots. $$ The difference is $$ 256(q^3 + 12 q^5 + 66 q^7 + 232 q^9 + 627 q^{11} + 1452 q^{13} + \cdots); $$ after a bit of experimentation we recognize this as $256q^3$ times the $12$-th power of $1+q^2+q^6+q^{12}+q^{20}+\cdots$, which is to say $256$ times the $12$-th power of the weight-$1/2$ modular form $\sum_{k=0}^\infty q^{(2k+1)^2/4}$. Such a formula, once surmised, can be proved by comparing initial segments of the $q$-expansions; I did this to $O(q^{61})$, which is way more than enough. The congruence mod $256$ follows because the $12$-th power of $1+q^2+q^6+q^{12}+q^{20}+\cdots$ clearly has integral coefficients.

Added later: This identity (and thus the congruence that Tito Piezas III asked for) gives a formula $(\sigma_5(n) - \rho(n)) / 256$ for the number of representations of $4n$ as the sum of $12$ odd squares, or equivalently of $(n-3)/2$ as the sum of $12$ triangular numbers. Following this lead, I soon found both the formula and the congruence in the paper

Ken Ono, Sinai Robins, and Patrick T. Wahl: On the representation of integers as sums of triangular numbers, Aequationes Math. 50 (1995) #1, 73-94

available on Ken Ono's page, where the enumeration is given (in the triangular-number form) as Theorem 7, and the congruence as a "simple consequence" of that theorem.

To keep the discussion alive and local, I add here further manifestations of the above behavior. Let $q = e^{2\pi i z}$,

III. 8th power

Define the numbers $a(n)$ according to

$$\begin{aligned}\eta(3z)^8 &= \sum_{n=1}^\infty a(n)q^n\\ &=q - 8q^4 + 20q^7 - 70q^{13}+64q^{16} +56q^{19} - 125q^{25} -160q^{28} + \dots\end{aligned}$$ Then I claim that $$a(n)\equiv \sigma_3(n) \mod 81.$$

IV. 6th power

Define the numbers $b(n)$ according to

$$\begin{aligned}\eta(4z)^6 &= \sum_{n=1}^\infty b(n)q^n\\ &=q - 6q^5 + 9q^9+10q^{13}-30q^{17} +11q^{25}+42q^{29}-70q^{37} + \dots\end{aligned}$$ Then I claim that $$b(n)\equiv \sigma_3(n) \mod 4.$$