A counterexample for $n = 4$ was given in the comments (2431, per @PeterTaylor). Monte Carlo simulation results of my own suggest counterexamples ought to increase with increasing $n$.

Moreover, the second paragraph of my question, about implications of the conjecture, is incorrect. The expected number of inversions for fixed $n$ is a constant, so it cannot covary with anything. Also, if a conjecture is strictly combinatorial, it only has implications for when $\rho = 0$. On reflection, I believe the relevant conjecture must be probabilistic, not combinatorial.

I therefore revise my conjecture on the relationship between fixed points and inversions in permutations as follows. The expected number of fixed points, $E(k)$, increases as the expected number of inversions, $E(j)$, decreases, for $n$ fixed and $\rho$ ranging over $[-1,1]$. This is very much the sort of conjecture one might prove (or disprove) via a Mallows tau model, as described in, e.g., Mukherjee (2016) and He (2022). They model a permutation as drawn from a distribution of permutations, with location parameter the identity permutation $(1, 2, ... , n)$, and spread parameter the expected Kendall tau distance. The latter is a rescaled Kendall's $\tau$ correlation coefficient.

In fact, I believe He (2022) in particular proves the revised conjecture for all $E(\tau) < 0$ (or $q < 1$, in his notation). He's Proposition 5.3 (p. 17) states that the expected number of fixed points in that case is equal to the probability that any one element in the permutation is a fixed point. Thus, expected fixed points range from $.5$ to $1$ as $\tau$ ranges from $-1$ to $0$. (The minimum expectation is $.5$ because, when the permutation is a complete inversion, it has no fixed points if $n$ is even but the middle element is always fixed for odd $n$, which averages to $.5$.)

His models for $E(\tau) > 0$ $(q > 1)$ are far more complex. He does not seem to explicitly derive the expected number of fixed points in that case. This is therefore only a partial answer. Any further answers or thoughts are welcome!

A counterexample for $n = 4$ was given in the comments (2431, per @PeterTaylor). Monte Carlo simulation results of my own suggest counterexamples ought to increase with increasing $n$.

Moreover, the second paragraph of my question, about implications of the conjecture, is incorrect. The expected number of inversions for fixed $n$ is a constant, so it cannot covary with anything. Also, if a conjecture is strictly combinatorial, it only has implications for when $\rho = 0$. On reflection, I believe the relevant conjecture must be probabilistic, not combinatorial.

I therefore revise my conjecture on the relationship between fixed points and inversions in permutations as follows. The expected number of fixed points, $E(k)$, increases as the expected number of inversions, $E(j)$, decreases, for $n$ fixed and $\rho$ ranging over $[-1,1]$. This is very much the sort of conjecture one might prove (or disprove) via a Mallows tau model, as described in, e.g., Mukherjee (2016) and He (2022). They model a permutation as drawn from a distribution of permutations, with location parameter the identity permutation $(1, 2, \dotsc , n)$, and spread parameter the expected Kendall tau distance. The latter is a rescaled Kendall's $\tau$ correlation coefficient.

In fact, I believe He (2022) in particular proves the revised conjecture for all $E(\tau) < 0$ (or $q < 1$, in his notation). He's Proposition 5.3 (p. 17) states that the expected number of fixed points in that case is equal to the probability that any one element in the permutation is a fixed point. Thus, expected fixed points range from $.5$ to $1$ as $\tau$ ranges from $-1$ to $0$. (The minimum expectation is $.5$ because, when the permutation is a complete inversion, it has no fixed points if $n$ is even but the middle element is always fixed for odd $n$, which averages to $.5$.)

His models for $E(\tau) > 0$ ($q > 1$) are far more complex. He does not seem to explicitly derive the expected number of fixed points in that case. This is therefore only a partial answer. Any further answers or thoughts are welcome!

A counterexample for $n = 4$ was given in the comments (2431, per @PeterTaylor). Monte Carlo simulation results of my own suggest counterexamples ought to increase with increasing $n$.

Moreover, the second paragraph of my question, about implications of the conjecture, is incorrect. The expected number of inversions for fixed $n$ is a constant, so it cannot covary with anything. Also, if a conjecture is strictly combinatorial, it only has implications for when $\rho = 0$. On reflection, I believe the relevant conjecture must be probabilistic, not combinatorial.

I therefore revise my conjecture on the relationship between fixed points and inversions in permutations as follows. The expected number of fixed points, $E(k)$, increases as the expected number of inversions, $E(j)$, decreases, for $n$ fixed and $\rho$ ranging over $[-1,1]$. This is very much the sort of conjecture one might prove (or disprove) via a Mallows tau model, as described in, e.g., Mukherjee (2016) and He (2022). They model a permutation as drawn from a distribution of permutations, with location parameter the identity permutation $(1, 2, ... , n)$, and spread parameter the expected Kendall tau distance. The latter is a rescaled Kendall's $\tau$ correlation coefficient.

In fact, I believe He (2022) in particular proves the revised conjecture for all $E(\tau) < 0$ (or $q < 1$, in his notation). He's Proposition 5.3 (p. 17) states that the expected number of fixed points in that case is equal to the probability that any one element in the permutation is a fixed point. Thus, expected fixed points range from $.5$ to $1$ as $\tau$ ranges from $-1$ to $0$. (The minimum expectation is $.5$ because, when the permutation is a complete inversion, it has no fixed points if $n$ is even but the middle element is always fixed for odd $n$, which averages to $.5$.)

His models for $E(\tau) > 0$ ($q > 1$) are far more complex. He does not seem to explicitly derive the expected number of fixed points in that case. This is therefore only a partial answer. Any further answers or thoughts are welcome!

A counterexample for $n = 4$ was given in the comments (2431, per @PeterTaylor). Monte Carlo simulation results of my own suggest counterexamples ought to increase with increasing $n$.

Moreover, the second paragraph of my question, about implications of the conjecture, is incorrect. The expected number of inversions for fixed $n$ is a constant, so it cannot covary with anything. Also, if a conjecture is strictly combinatorial, it only has implications for when $\rho = 0$. On reflection, I believe the relevant conjecture must be probabilistic, not combinatorial.

I therefore revise my conjecture on the relationship between fixed points and inversions in permutations as follows. The expected number of fixed points, $E(k)$, increases as the expected number of inversions, $E(j)$, decreases, for $n$ fixed and $\rho$ ranging over $[-1,1]$. This is very much the sort of conjecture one might prove (or disprove) via a Mallows tau model, as described in, e.g., Mukherjee (2016) and He (2022). They model a permutation as drawn from a distribution of permutations, with location parameter the identity permutation $(1, 2, ... , n)$, and spread parameter the expected Kendall tau distance. The latter is a rescaled Kendall's $\tau$ correlation coefficient.

In fact, I believe He (2022) in particular proves the revised conjecture for all $E(\tau) < 0$ (or $q < 1$, in his notation). He's Proposition 5.3 (p. 17) states that the expected number of fixed points in that case is equal to the probability that any one element in the permutation is a fixed point. Thus, expected fixed points range from $.5$ to $1$ as $\tau$ ranges from $-1$ to $0$. (The minimum expectation is $.5$ because, when the permutation is a complete inversion, it has no fixed points if $n$ is even but the middle element is always fixed for odd $n$, which averages to $.5$.)

His models for $E(\tau) > 0$ $(q > 1)$ are far more complex. He does not seem to explicitly derive the expected number of fixed points in that case. This is therefore only a partial answer. Any further answers or thoughts are welcome!