Ramanujan's tau function, $691$ congruence, and $\eta(z)^{12}$ Let $q = e^{2\pi i\,z}$.

I. 24th power

The Ramanujan tau function $\tau(n)$ is given by the expansion of the Dedekind eta function $\eta(z)$'s $\text{24th}$ power. Then
$$\begin{aligned}\eta(z)^{24} &= \sum_{n=1}^\infty\tau(n)q^n\\&=q - 24q^2 + 252q^3 - 1472q^4 + 4830q^5 - 6048q^6 - 16744q^7 + \dots\end{aligned}$$
Ramanujan observed that 
$$\tau(n)\equiv\sigma_{11}(n)\ \bmod\ 691\tag1$$

II. 12th power

Assume the rho function $\rho(n)$ as,
$$\begin{aligned}\eta(2z)^{12} &= \sum_{n=1}^\infty\rho(n)q^n\\&=q - 12q^3 + 54q^5 - 88q^7 -99q^9 +540q^{11} - 418q^{13} -648q^{15} + \dots\end{aligned}$$
Note the odd powers. Is it true that
$$\rho(n)\equiv\sigma_{5}(n)\ \bmod\ 2^8\tag2$$
analogous to $(1)$? 
P.S. It's true for the first 10000 coefficients in OEIS A000735.
 A: 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
$$b(n)\equiv \sigma_3(n) \mod 4.$$
A: 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.
