Define the polynomials
$$\begin{align*} Q_0(x) &= b_0, \\ Q_1(x) &= b_1 (x + 1), \\ Q_2(x) &= b_2 (x + 1)(x + 3), \\ Q_3(x) &= b_3 (x + 1)(x + 3)(x + 5), \\ Q_4(x) &= b_4 (x + 1)(x + 3)(x + 5)(x + 7), \\ Q_5(x) &= b_5 (x + 1)(x + 3)(x + 5)(x + 7)(x + 9), \\ Q_6(x) &= b_6 (x + 1)(x + 3)(x + 5)(x + 7)(x + 9)(x + 11), \end{align*}$$
where
$$b_0 = 2^{10}, \quad b_1 = 2^9, \quad \dots, b_2 = 2^7, \quad b_3 = 2^6, \quad b_4 = 2^3, \quad b_5 = 2^2, \quad b_6 = 2^0.$$
The product of $i$ consecutive even integers is divisible by $2^i \cdot i!$. Therefore, for $i = 0, 1, 2, 3, 4, 5, 6$, we obtain that the product of $i$ consecutive even integers is divisible by $2^0, 2^1, 2^3, 2^4, 2^7, 2^8, 2^{10}$, respectively. This implies that, for any odd integer $x$ and $i = 0, \dots, 6$, $Q_i(x)$ is divisible by $2^{10}$.
A polynomial $P(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5$ with integer coefficients is called reduced if, for $i = 0, \dots, 5$,
$$0 \le a_i < b_i. \qquad (\dagger)$$
Clearly, there are exactly $b_0b_1\dots b_5 = 2^{10+9+7+6+3+2} = 2^{37}$ distinct reduced polynomials.
We show that, for every polynomial $P(x)$ with integer coefficients, there exists a reduced polynomial $\bar{P}(x)$ such that $P(x)$ and $\bar{P}(x)$ have the same remainder sequence.
First note that, for $i = 0, \dots, 6$, and any polynomial $R(x)$ with integer coefficients $P(x)$ and $P(x) - R(x)Q_i(x)$ have the same remainder sequence. This follows from the fact that $Q_i(x)$ is divisible by $2^{10}$, for any odd integer $x$.

If the degree $d$ of $P(x) = a_0 + \cdots + a_d x^d$ is higher than $5$ we may replace $P(x)$ by $P(x) - a_d x^{d-6} Q_6(x)$. Indeed, the polynomial $P(x) - a_d x^{d-6} Q_6(x)$ has smaller degree than $P(x)$ and has the same remainder sequence as $P(x)$. We may continue this until we obtain a polynomial that of degree at most $5$ that has the same remainder sequence as $P(x)$.
We assume now that $P(x)$ has degree no higher than $5$. If $P(x)$ is reduced we are done. Otherwise, let $i$ be the highest degree of a coefficient $a_i$ of $x^i$ that does not satisfy the range condition $(\dagger)$. If $q$ is the quotient obtained by dividing $a_i$ by $b_i$ then $P(x)$ and $P(x) - qQ_i(x)$ have the same remainder sequence and the coefficient at degree $i$ in $P(x) - qQ_i(x)$ is in the correct range $0, \dots, b_i - 1$.
We repeat this procedure with the next highest degree that has a coefficient out of range until we reach a reduced polynomial that has the same remainder sequence as $P(x)$.

We now consider the $2^{37}$ reduced polynomials.
Let $a = 2^9+1$ and $b = 1$. Then $P(a)-P(b) = (a-b)(a_1+a_2A_2+a_3A_3+a_4A_4+a_5A_5)$, where $A_2 = a+b$, $A_3 = a^2+ab+b^2$, $A_4 = a^3+a^2b+ab^2+b^3$ and $A_5 = a^4+a^3b+a^2b^2+ab^3+b^4$. Since both $a$ and $b$ are odd, $A_2$ and $A_4$ are even, $A_3$ and $A_5$ are odd, and the parity of $a_1+a_2A_2+a_3A_3+a_4A_4+a_5A_5$ is the same as the parity of $a_1+a_3+a_5$. Therefore, if $a_1+a_3+a_5$ is even $P(a)-P(b)$ is divisible by $2^{10}$ and the sequence of remainders of $P(x)$ is not a permutation.
For an odd integer $x$, the parity of $P(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5$ is the same as the parity of the sum $a_0 + a_1 + \cdots + a_5$. Thus, only polynomials with odd sum of coefficients have odd remainders.
Therefore, there are no more remainder sequences that are permutations of $1, 3, \dots, 1023$ than there are reduced polynomials $P(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5$ for which both $a_1 + a_3 + a_5$ and $a_0 + a_1 + a_2 + a_3 + a_4 + a_5$ are odd.
There are exactly $2^{36}$ reduced polynomials for which $a_1+a_3+a_5$ is odd. This can be seen by pairing up every reduced polynomial $P(x)$ in which $a_1$ is even with the polynomial $P(x)+x$. Exactly one of the two polynomials in each such pair has odd sum $a_1+a_3+a_5$.
There are exactly $2^{35}$ reduced polynomials for which both $a_1+a_3+a_5$ and $a_0+a_1+a_2+a_3+a_4+a_5$ are odd. This can be seen by pairing up every reduced polynomial $P(x)$ in which $a_1+a_3+a_5$ is odd and $a_0$ is even with the polynomial $P(x)+1$. Exactly one of the two polynomials in each such pair has odd sum $a_0+a_1+a_2+a_3+a_4+a_5$.