You're confused in a few ways. :-)
First, Fiddler has always behaved like this.
Second, I'm betting that you got your Android to send traffic to Fiddler using iptables or some other hack on the device? The problem with doing that is that the request sent off the device doesn't have a proper Host in the CONNECT request; it instead uses the IP address of the target. By default, Fiddler presents a certificate bearing the CN = the host that it was asked to CONNECT to. Since the iptables hack doesn't provide Fiddler with that hostname, you get the name mismatch.
Now, there are two name mismatches to consider: 1: Fiddler's displayed error, which it shows because the server didn't present a certificate that matched the Host of the request. 2: The Android displayed error, which it shows because Fiddler presented a certificate that didn't match the host that the Android browser thought it was making.
Fixing #1 is pretty simple, you can just ignore the warning.
For #2, fortunately, someone else reported this issue to me a while ago, and so you have some options depending on which Android build you're using.
If the Android build is current enough, you can set the fiddler.network.https.SetCNFromSNI preference to True, which will cause Fiddler to generate the SubjectCN using the server name indicator from the client's TLS handshake instead of using the HOST field from the CONNECT. If the client is properly using SNI, you won't get any errors.
If the client isn't sending SNI, you can fix mismatches on a one-off basis:
if (oSession.HTTPMethodIs("CONNECT") && oSession.HostnameIs("74.125.71.84")){
oSession.oFlags["x-OverrideCertCN"] =
"hostnameToUseInCertificate.com";
}
Now, some folks wonder: "Why can't Fiddler just use whatever CN the server used?" The answer is that Fiddler's handshake with the client occurs before the handshake with the server, because the client's CONNECT request might be for a WebSocket tunnel rather than a HTTPS tunnel, and if Fiddler were to blindly HTTPS handshake with the server before handshaking with the client, the WebSocket scenario would be broken.