This is the second part of the answer. Suppose $E^n$ is a flat torus admitting a conformal self-map $\varphi_d$ of degree $d$ for every $d=1,2,3,\ldots$. We prove that this is only possible when $n=1$.

Algebraic reformulation: Fix a positive definite symmetric bilinear form $Q$ on $\mathbb{R}^n$, $n\geq 2$. Call an integer $n\times n$ matrix $M$ conformal if $M^t Q M$ is a positive real multiple of $Q$. Degree of such a matrix is $\det M$. We prove that it is not possible to have a conformal matrix of degree $d$ for each $d=1,2,3,\ldots$. Suppose the contrary, i.e. there is such an integer matrix $M_d$ for every $d$.

First, by taking determinants  for every $d$ we find the coefficient of proportionality
$$
M_d^t Q M_d= d^{2/n} Q.
$$
Normalize $Q$ so that $Q_{11}=1$. For any vectors $u,v$ denote $(u,v)=u^t Q v$. Let $v_d$ be the first column of $M_d$. Then we have
$$
(v_d, v_d) = d^{2/n}\qquad (d=1,2,3,\ldots)
$$
We claim that this is impossible. Consider the case $n=2$ first. Note that no two among $v_1, v_2, v_3$ can be collinear. Hence $v_3=\alpha v_1 + \beta v_2$ for some $\alpha, \beta\in\mathbb{Q}$. This allows to compute all the entries of $Q$ out of $\alpha, \beta$ and deduce that they are rational. So we have $a,b,c\in\mathbb{Q}$ so that the equation $a x^2 + b xy + c y^2=d$ has solutions in integers for every $d$, but $b^2-4a c<0$. This impossible: by Chebotarev density theorem one can choose a prime $p$ such that $p$ doesn't divide the numerators and the denominators of $a,b,c$ and the equation $a x^2 + b x + c=0$ has no roots mod $p$. Setting $d=p$ leads to a contradiction.

Now consider the case $n\geq 3$. Consider the numbers of the form $p^{2/n}$ for prime numbers $p>n$. They are linearly independent over $\mathbb{Q}$ because the field extension generated by $p^{2/n}$ is ramified at $p$, and can only be further ramified at the divisors of $n$. On the other hand, consider the sequence of integer $n\times n$ matrices $v_p v_p^t$. We have an infinite sequence of elements of a finite dimensional vector space, so there must be a linear relation
$$
\sum_{i=1}^N c_i v_{p_i} v_{p_i}^t = 0 \qquad ((c_1,c_2,\ldots,c_N)\in\mathbb{Q}^N\setminus \{0\})
$$
for prime numbers $p_1,p_2,\ldots,p_N>n$. This implies
$$
\sum_{i=1}^N c_i p_i^{2/n} = \sum_{i=1}^N c_i (v_{p_i}, v_{p_i}) = \sum_{i=1}^N c_i \operatorname{trace}(Q v_{p_i} v_{p_i}^t)=0,
$$
a contradiction.