In polynomial functions, is the midpoint of two adjacent minimum and maximum always a point of inflection?
You are probably only trying cubics. If you have the function $f(x)=ax^3+bx^2+cx+d$ the local extrema come at the roots of $3ax^2+2bx+c=0$, which are $\frac {-b\pm \sqrt{b^2-3ac}}{3a}$. The midpoint of these is $\frac {-b}{3a}$. The inflection point is at the root of $6ax+2b=0$, which is $\frac {-b}{3a}$
If you try higher degree polynomials it fails. Take $f(x)=x^4-2x^2+1$ The extrema are at roots of $4x^3-4x=0$, so at $0,\pm 1$. The inflection points are at the roots of $12x^2-4=0$, which are $\pm \frac 1{\sqrt 3}$, not $\pm \frac12$
In general, it is not.
$$x^4-2x^2$$ has a maximum at $x=0$ and a minimum $x=1$ and an inflection point at $x=\dfrac1{\sqrt3}.$
Having a de-centered inflection is tantamount to having a de-centered extremum between two roots in the first derivative, which is also quite possible.
Consider the function
$$y'(x)=(x-a)(x-b)(x-d).$$
If has an extremum where
$$(x-a)(x-b)+(x-a)(x-d)+(x-b)(x-d)=0.$$
If we want this to occur at $x=c$, we can set
$$d=c+\frac{(c-a)(c-b)}{2c-a-b}.$$
So the antiderivative of
$$(x-a)(x-b)\left(x-c-\frac{(c-a)(c-b)}{2c-a-b}\right)$$
has two extrema and an inflection point wherever you want (except in the middle !)
This is true for cubic polynomials $p$: If $p$ has local extrema, then the quadratic function $p'$ has roots and these roots are symmetric about $x = r$, where $r$ is the unique root of $p''$ (so that $(r, p(r))$ is an inflection point). Of course, the claim is vacuous for linear and quadratic polynomials.
On the other hand, the claim is in general false for polynomials of degree $> 3$: If a quartic polynomial has more than one extremum, it has three, and so $p'$ has three real roots. By translating, dilating, and reflecting (all of which preserve midpoints) we may as well assume that $$p'(x) = x (x - r) (x - 1),$$ so that the extrema of $p$ occur at $0, r, 1$, where $0 < r < 1$. At the midpoints $\frac{1}{2} r$, $\frac{1}{2} (r + 1)$ of consecutive roots, we compute that $$p''\left(\tfrac{1}{2}r\right) = -\tfrac{1}{4} r^2 \qquad \textrm{and} \qquad p''\left(\tfrac{1}{2}(r + 1)\right) = -\tfrac{1}{4} (1 - r)^2,$$ which by hypothesis are never zero, so these are not inflection points. Thus, we can conclude for quartic polynomials that the claim never holds (at least in the nonvacuous case, that is, when the polynomial has multiple extreme).
The simple example $p(x) = x^5 - x$ shows that the claim can hold for specific quintic polynomials, but it fails in general in the degree $5$ case.