$\newcommand{\la}{\lambda}\newcommand{\de}{\delta}\newcommand{\ep}{\delta}$The moment of inertia of $C$ about the line $l_{p,a}$ through a point $p=(p_1,p_2)$ orthogonal to a unit vector $a=(a_1,a_2)$ is \begin{equation*} M(p,a):=\int_C(a\cdot(x-p))^2\,dx, \end{equation*} where $\cdot$ denotes the dot product. So, \begin{equation*} M(p,a)=a\cdot G(p)a, \end{equation*} where $G(p)$ is the $2\times2$ matrix with entries \begin{equation*} G(p)_{i,j}=\int_C(x_i-p_i)(x_j-p_j)\,dx \\ =\int_C x_i x_j\,dx+p_i p_j \int_C\,dx-p_i\int_C x_j\,dx-p_j\int_C x_i\,dx. \end{equation*} By shifting and rescaling, without loss of generality (wlog) $\int_C\,dx=1$ and $\int_C x_j\,dx=0$ for $j=1,2$. So, \begin{equation*} G=H+p\otimes p, \end{equation*} where $H$ is the $2\times2$ matrix with entries $H_{ij}:=\int_C x_i x_j$ and $p\otimes p$ is the $2\times2$ matrix with entries $p_i p_j$.
The matrix $H$ is positive definite. So, by a rotation, wlog $H$ is a diagonal positive-definite matrix, so that \begin{equation*} M(p,a)=\la_1 a_1^2+\la_2 a_2^2+(p_1 a_1+p_2 a_2)^2, \end{equation*} where $\la_1$ and $\la_2$ are real numbers such that, wlog, $\la_1\ge\la_2>0$.
Fix now a point $p$ and a unit vector $a$, and let \begin{equation*} m:=M(p,a). \end{equation*} Suppose now that the moment of inertia of $C$ about the line $l_{p,b}$ (through $p$ orthogonal to a unit vector $b=(b_1,b_2)$) is $m$ as as well. This means that $b$ is a solution to the system of equations \begin{equation*} \la_1 b_1^2+\la_2 b_2^2+(p_1 b_1+p_2 b_2)^2=m,\quad b_1^2+b_2^2=1. \tag{1}\label{1} \end{equation*} Let us say that a unit vector $b$ is good and that the line $l_{p,b}$ is good if $b$ is a solution of system \eqref{1}. Given $p$, the good lines $l_{p,b}$ are in the one-to-one correspondence with the sets of the form $\{b,-b\}$, where $b$ is good. Say that unit vector $b$ and $-b$ are equivalent to each other.
The first equation in \eqref{1} is the equation of an ellipse centered at the origin and the second equation in \eqref{1} is the equation of the unit circle. So, we have either
- Case 1: the ellipse is the same as the circle, and then any line through $p$ is good or
- Case 2: the ellipse is not the same as the circle, and then we only have two non-equivalent good unit vector $b$'s, and thus only two good lines.
Clearly, Case 1 occurs only if $p_1 p_2=0$. So, given the condition $\la_1\ge\la_2>0$, Case 1 occurs only if $p_1=0$ and $p_2^2=\la_1-\la_2$.
However, then both corresponding points $(p_1,p_2)=(0,\pm\sqrt{\la_1-\la_2})$ may be not in $C$. E.g., for small $\de\in(0,1)$, let $C=[-\frac1{2\de},\frac1{2\de}]\times[-\frac\de2,\frac\de2]$. Then $C$ is convex with the centroid at the origin and with area $1$, the matrix $H$ is diagonal with diagonal entries $\la_1=\frac1{12\ep^2}$ and $\la_2=\frac{\ep^2}{12}$, so that the points $(0,\pm\sqrt{\la_1-\la_2})=(0,\sqrt{1/12}\sqrt{1-\ep^4}/\ep)$ are not in this $C$ if $\de$ is small enough.
It is also clear from the above consideration that your Lemma will not hold in general (if, say, $p=0$ and $\la_1>\la_2$.)