Rephrase in terms of graph theory:
Let $G$ be a connected graph on $50$ vertices. Given a vertex $A$, a permutation $(V_1, \dots, V_{50})$ of the vertices of $G$ is called an antitour from $A$ if the following condition holds for every integer $k$ with $1 \le k \le 50$: there is no walk of length $k$ from $A$ to $V_k$. Given that every vertex in $G$ admits at least one antitour, and that $G$ has a Hamiltonian path, find the least possible number of antitours from some vertex of $G$.
Let $N = 50$; we claim that the answer is $((N/2)!)^2$.

Bound. Observe that if $V$ can be reached by a walk of length $d$, then it can be reached by a walk of length $d + 2k$ for any non-negative integer $k$, simply by going to $V$ and then going back-and-forth on some adjacent edge $k$ times.

Claim 1 The graph is bipartite.
Proof. Indeed, suppose there is some odd cycle $A_0, A_1, \dots, A_{2i}$. Consider an antitour $C_1, \dots, C_N$ from $A_0$. Note that for $1 \le k \le 2i$, $A_k$ can be reached from $A_0$ by $k$ moves (and thus in $k+2j$ moves) and also in $2i+1-k$ (and thus in $2i+1-k+2j$ moves). Now any integer $n \ge 2i$ can be written as either $k+2j$ (if $n \equiv k \pmod 2$) or as $2i+1-k+2j$ (if $n \not\equiv k \pmod 2$), so none of the $A_k$'s can be $C_n$ for $n \ge 2i$. Thus the vertices $A_1, \dots, A_{2i}$ must be $C_1, \dots, C_{2i-1}$ in some order, which is impossible. $\square$

Proof. Alternately, one can also argue the initial part as is and then since the distance of any vertex not in the cycle is at most $N - (2i+1)$. Now, from any vertex $A_j$ in the cycle, consider an anti-tour. Let the $N$th in this anti-tour be $V_j$. Now, the distance of $V_j$ to the cycle is at most $N - (2i+1)$ and there is a path of length $\le 2i+1$ from $A_j$ to the vertex to which $V_j$ has the shortest path in the cycle with odd or even number of steps. Thus, there is an even and an odd path from $A_j$ to $V_j$ in $\le N$ moves. This is clearly a contradiction! $\square$

Since the Hamiltonian path must contain an equal number of vertices from each part, the parts have $N/2$ vertices each. If the capital is in part $A = \{A_1, \dots, A_{N/2}\}$, and the other part is $B = \{B_1, \dots, B_{N/2}\}$, then one can form an antitour by putting the $A_i$'s in the odd positions in any order and $B_i$'s in the even positions in any order. There are $((N/2)!)^2$ ways to do this, which proves our bound.

Construction. Consider the complete bipartite graph $K_{N/2,N/2}$, with parts $\{A_1, \dots, A_{N/2}\}$ and $\{B_1, \dots, B_{N/2}\}$. Any antitour from $A_1$ must have all the $B_i$'s in the even positions since any $B_i$ is reachable in $1$ move (and thus in any odd number of moves). Thus there are at most $((N/2)!)^2$ antitours for this. $\square$

We can have alternate constructions. For example: Let $A_1, \dots, A_{50}$ be the Hamiltonian path, with $A_1$ capital. Add the edges $A_1A_4, A_1A_6, \dots, A_1A_{50}$ and $A_2A_5, A_2A_7, \dots, A_2A_{49}$.