Suppose we have a locally compact group $G$ and a closed unimodular normal subgroup $N$.
Let $\mu$ be a Borel probability measure on $G$ which is left $N$-invariant.
Does it follow that $\mu$ is right $N$-invariant?
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Yes. This is true in the particular case when $\mu$ is a left Haar measure (i.e. is left $G$-invariant) — without assuming $\mu$ finite.
See Corollary B.1.7 in the book "Kazhdan's Property T" by Bekka-Harpe-Valette (link at Bekka's page). In this setting, that $N$ is normal ensures that $G/N$ has a [Radon] invariant Borel measure, and since $N$ unimodular, the conclusion of the corollary is that the modular map $\Delta_G$ is trivial on $N$, which exactly means that the left Haar measure $\mu=\mu_G$ is right $N$-invariant.
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$\begingroup$ I'm not sure that I follow this argument. It seems that we are assuming $\mu$ is the Haar measure on $G$? But I am mostly interested in the cases where this is not so. $\endgroup$– KimCommented Jan 23 at 4:20
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$\begingroup$ @Kim Indeed, I misread the question. I edited to say what I'm actually proving. Maybe one could then try the case when assuming that $\mu$ has density with respect to the Haar measure. $\endgroup$– YCorCommented Jan 23 at 7:52
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$\begingroup$ What about the case $G$ is compact and $N=G$? Does it follow that the measure $\mu$ necessarily the Haar measure? (since you don't assume $\mu$ to be outer or inner regular). $\endgroup$– user515519Commented Jan 23 at 16:02