= Solution
There is an <orientation> convention in this statement: the two <volume forms> must be positive for one fixed <orientation>, as is usual in <Moser theorem for volume forms>. Without that convention, the stated necessity is false. For example, on an oriented <two-sphere>, take a positive <volume form> $\beta_0$ and an <orientation-reversing diffeomorphism> $r$, and set $\beta_1=r^*\beta_0$. Then $(r^{-1})^*\beta_1=\beta_0$ but $\int\beta_1=-\int\beta_0$. We prove the intended result with a common <orientation>.
If $\phi^*\beta_1=\beta_0$, positivity forces $\phi$ to preserve the <orientation>, and integration of the <pullback of a differential form> gives equality of the integrals.
For sufficiency, <top-degree de Rham cohomology> gives a <differential form> $\alpha$ of degree $n-1$ with $\beta_1-\beta_0=d\alpha$. Put $\beta_t=(1-t)\beta_0+t\beta_1$. Positivity makes every $\beta_t$ a <volume form>. The <interior product of a differential form> gives an <linear isomorphism>
$$
T_pM\longrightarrow\Lambda^{n-1}T_p^*M,\qquad v\longmapsto\iota_v\beta_t.
$$
Thus there is a unique smooth time-dependent <vector field> $X_t$ satisfying $\iota_{X_t}\beta_t=-\alpha$. Since $M$ is <compact>, its <local flow> exists throughout $0\leq t\leq1$ and defines a <smooth isotopy> $\phi_t$ starting at the identity. By <Cartan's magic formula>, and since a top-degree <differential form> has zero <exterior derivative>,
$$
\frac{d}{dt}\phi_t^*\beta_t
=\phi_t^*\bigl(d\alpha+\mathcal L_{X_t}\beta_t\bigr)
=\phi_t^*\bigl(d\alpha+d\iota_{X_t}\beta_t\bigr)=0.
$$
Hence
$$
\boxed{\phi_1^*\beta_1=\beta_0.}
$$
The zero-dimensional case, a single point, is immediate from equality of the integrals.
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