Solution 1. One such integer is $n = 91$. We show there is no anchor $m > 1$ such that $m \times n$ is also an anchor for this value of $n$. Let $m > 1$ be an anchor. We consider separately the cases $3 \le m \le 9$ and $m \ge 11$.
For small anchors $m$, we have $3 \times 91 = 273$, $5 \times 91 = 455$, $7 \times 91 = 637$ and $9 \times 91 = 819$, none of which are anchors.

When $m \ge 11$, let the last two digits of $m$ be $m_1$ and $m_0$, with $m_0, m_1 \in \{1, 3, 5, 7, 9\}$. Then we can write (for some integer $k \ge 0$): $m = 100k + 10m_1 + m_0$ so that:
$$\begin{aligned} 91m &= (90 + 1)(100k + 10m_1 + m_0) \\ &= 100 \times (91k + 9m_1) + 10(9m_0 + m_1) + m_0 \end{aligned}$$
The penultimate digit is then $9m_0 + m_1 \pmod{10}$ which is even, and so $91m$ cannot be an anchor.

Solution 2. One such integer is $n = 99$. We show there is no anchor $m > 1$ such that $m \times n$ is also an anchor for this value of $n$. Let $m > 1$ be an anchor. We consider separately the cases $3 \le m \le 9$ and $m \ge 11$.
For small anchors $m$, we have $3 \times 99 = 297$, $5 \times 99 = 495$, $7 \times 99 = 693$ and $9 \times 99 = 891$, none of which are anchors.

When $m \ge 11$, let the last two digits of $m$ be $m_1$ and $m_0$, with $m_0, m_1 \in \{1, 3, 5, 7, 9\}$. Then we can write (for some integer $k \ge 0$): $m = 100k + 10m_1 + m_0$ so that
$$\begin{aligned} 99(100k + 10m_1 + m_0) &\equiv -(10m_1 + m_0) \equiv 100 - 10m_1 - m_0 \\ &\equiv 10(9 - m_1) + (10 - m_0) \pmod{100}. \end{aligned}$$
This shows that the second last digit of $99m$ is equal to $9 - m_1$, and this is even since $m_1$ is odd. Hence, $99m$ is not an anchor for any anchor $m > 1$.

Solution 3. One such integer is $n = 911$. We show there is no anchor $m > 1$ such that $m \times n$ is also an anchor for this value of $n$. The argument below applies to any $n = 911 \dots 11 > 91$.

First note that $9a$, for any number $1 < a < 11$, has two digits, the sum of which is divisible by 9 and smaller than 18, hence equal to 9. The units digit of $9a$ is odd if $a$ is odd, hence the leading digit is equal to the even number $9 - a$. In particular, the leading digit of $911 \cdot m$ is even for $m = 3, 5, 7, 9$, because there are no carries from the multiplications by 1.
If $m > 10$ is an anchor, we have $m \equiv 10a + b \pmod{100}$ with $a, b$ being odd digits. Hence $911m \equiv 11(10a + b) \equiv 10(a + b) + b \pmod{100}$ which means that the units digit of $911m$ is $b$ and the hundreds digit is equal to $a + b$ or $a + b - 10$, both of which are even.

Solution 4. It suffices to find a single anchor $n$ such that $m \times n$ is a non-anchor for all anchors $m > 1$. There are many such $n$. Here are some cases where $n \equiv 11 \pmod{20}$ with manual checks for $m = 3, 5, 7, 9$: